Q.Show that the three lines with direction cosines $\frac{12}{13}, \frac{-3}{13}, \frac{-4}{13}$; $\frac{4}{13}, \frac{12}{13}, \frac{3}{13}$; $\frac{3}{13}, \frac{-4}{13}, \frac{12}{13}$ are mutually perpendicular.

Showing the 12 most recent of 21 on this concept.

1. TG EAPCET 2022Set eng-2022-07-20-FN1 markMCQ
   Q.If $ax^2 + 6xy - 2y^2 = 0$ represents a pair of perpendicular lines and $9x^2 + 2hxy + 4y^2 = 0$ $(h > 0)$ represents a pair of coincident lines then $h =$ (A) $3a$ (B) $2a$ (C) $a$ (D) $4a$

   ›Reveal solutionSolution

   Apply the condition for perpendicular lines (the sum of the $x^2$ and $y^2$ coefficients is zero) to the first equation to find $a$, and the condition for coincident lines (discriminant zero) to the second equation to find $h$. Comparing the two values gives $h = 3a$, which is option (A).

   We have two homogeneous second-degree equations, each representing a pair of straight lines through the origin. The first represents perpendicular lines; the second, coincident lines. Each condition translates into a specific relation among the coefficients.

   For a general homogeneous pair of lines $Ax^2 + 2Hxy + By^2 = 0$:

   • The lines are perpendicular if $A + B = 0$.
   • The lines are coincident if $H^2 - AB = 0$.

   1. First equation: $ax^2 + 6xy - 2y^2 = 0$. Comparing with the general form, $A = a$, $2H = 6 \Rightarrow H = 3$, and $B = -2$. For perpendicular lines:

   $a + (-2) = 0 \quad\Rightarrow\quad a = 2.$

   2. Second equation: $9x^2 + 2hxy + 4y^2 = 0$, with $h > 0$. Here $A = 9$, $2H = 2h \Rightarrow H = h$, and $B = 4$. For coincident lines:

   $h^2 - (9)(4) = 0 \quad\Rightarrow\quad h^2 = 36 \quad\Rightarrow\quad h = \pm 6.$

   Since $h > 0$, we take $h = 6$.

   3. Relating $h$ and $a$: we found $a = 2$ and $h = 6$, so $h = 3a$ (since $3 \times 2 = 6$). This matches option (A).

   Watch out

   A common mistake is to forget that the coefficient of $xy$ in the standard form is $2H$, not $H$. In the first equation, $6xy$ means $2H = 6$, so $H = 3$, not $6$. Similarly, in the second equation, $2hxy$ means $2H = 2h$, so $H = h$.

   Tip

   For perpendicular lines through the origin, the condition $A + B = 0$ is quick and avoids solving for slopes. For coincident lines, the discriminant condition $H^2 - AB = 0$ is the only condition needed.

   ✓Final answer

   The value is $h = 3a$, which corresponds to option (A).

2. TG EAPCET 2025Set eng-2025-05-02-AN1 markMCQ
   Q.The value of ‘a’ for which the equation $(a^2-3)x^2+16xy-2ay^2+4x-8y-2=0$ represents a pair of perpendicular lines is (A) $2$ (B) $-1$ (C) $3$ (D) $4$

   ›Reveal solutionSolution

   For a second-degree equation to represent a pair of perpendicular lines, the sum of the coefficients of $x^2$ and $y^2$ must be zero. Here, that condition gives $a = -1$, which matches option (B).

   ---

   The key idea: A general second-degree equation $Ax^2 + 2Hxy + By^2 + 2Gx + 2Fy + C = 0$ represents a pair of straight lines if its determinant (the "discriminant" of the conic) vanishes. If those lines are also perpendicular, then the coefficients of $x^2$ and $y^2$ satisfy $A + B = 0$. Why? Because for two perpendicular lines with slopes $m_1$ and $m_2$, we have $m_1 m_2 = -1$. In the pair-of-lines form, the product of slopes equals $A/B$ (when the lines are not vertical/horizontal), so perpendicularity implies $A/B = -1$, i.e., $A + B = 0$. This is a clean, direct condition — no need to find the actual lines.

   Let’s apply it step by step.

   1. Identify the coefficients from the given equation:

   $(a^2 - 3)x^2 + 16xy - 2a y^2 + 4x - 8y - 2 = 0$

   Compare with the standard form $Ax^2 + 2Hxy + By^2 + 2Gx + 2Fy + C = 0$:

   • $A = a^2 - 3$
   • $2H = 16 \Rightarrow H = 8$
   • $B = -2a$
   • $2G = 4 \Rightarrow G = 2$
   • $2F = -8 \Rightarrow F = -4$
   • $C = -2$

   2. Apply the perpendicularity condition: For perpendicular lines, we require

   $A + B = 0$

   So:

   $(a^2 - 3) + (-2a) = 0$

   Simplify:

   $a^2 - 2a - 3 = 0$

   3. Solve the quadratic:

   $a^2 - 2a - 3 = (a - 3)(a + 1) = 0$

   Hence $a = 3$ or $a = -1$.

   4. Check which also satisfies the “pair of lines” condition (the determinant must vanish). The equation must actually represent two lines, not just satisfy perpendicularity hypothetically. Compute the determinant condition:

   $\begin{vmatrix} A & H & G \\ H & B & F \\ G & F & C \end{vmatrix} = 0$

   Substitute:

   $\begin{vmatrix} a^2 - 3 & 8 & 2 \\ 8 & -2a & -4 \\ 2 & -4 & -2 \end{vmatrix} = 0$

   • For $a = 3$: $A = 6$, $B = -6$, matrix becomes:

   $\begin{vmatrix} 6 & 8 & 2 \\ 8 & -6 & -4 \\ 2 & -4 & -2 \end{vmatrix}$

    Compute: $6(12 - 16) - 8(-16 + 8) + 2(-32 + 12) = 6(-4) - 8(-8) + 2(-20) = -24 + 64 - 40 = 0$. So it works.
   

   • For $a = -1$: $A = (-1)^2 - 3 = -2$, $B = -2(-1) = 2$, matrix:

   $\begin{vmatrix} -2 & 8 & 2 \\ 8 & 2 & -4 \\ 2 & -4 & -2 \end{vmatrix}$

    Compute: $-2(-4 - 16) - 8(-16 + 8) + 2(-32 - 4) = -2(-20) - 8(-8) + 2(-36) = 40 + 64 - 72 = 32 \neq 0$.  
    So $a = -1$ does **not** yield a pair of lines — the determinant is non-zero, meaning it represents a hyperbola or other conic, not two straight lines.
   

   Watch out

   A common mistake is to stop at $A + B = 0$ and pick both roots. But the equation must also represent a pair of lines (the conic must be degenerate). Always verify the determinant condition.

   Thus only $a = 3$ satisfies both conditions.

   ✓Final answer

   The correct option is (C).

   ANSWER: C

3. TG EAPCET 2025Set eng-2025-05-04-AN1 markMCQ
   Q.If the circle $S = 0$ intersect the three circles $S_1 \equiv x^2 + y^2 + 4x - 7 = 0$, $S_2 \equiv x^2 + y^2 + y = 0$ and $S_3 \equiv x^2 + y^2 + \frac{3}{2}x + \frac{5}{2}y - \frac{9}{2} = 0$ orthogonally, then radical axis of $S = 0$ and $S_1 = 0$ is (A) $4x - y - 7 = 0$ (B) $x + y - 3 = 0$ (C) $4x + y - 3 = 0$ (D) $x - y - 2 = 0$

   ›Reveal solutionSolution

   The radical axis of $S=0$ and $S_1=0$ is $4x + y - 3 = 0$ — option (C).

   Let $S \equiv x^2+y^2+2gx+2fy+c=0$. The orthogonality condition with a circle $x^2+y^2+2g_ix+2f_iy+c_i=0$ is $2gg_i+2ff_i=c+c_i$.

   With $S_1\equiv x^2+y^2+4x-7=0$ $(g_1=2,f_1=0,c_1=-7)$:

   $4g=c-7.$

   With $S_2\equiv x^2+y^2+y=0$ $(g_2=0,f_2=\tfrac12,c_2=0)$:

   $f=c.$

   With $S_3\equiv x^2+y^2+\tfrac32x+\tfrac52y-\tfrac92=0$ $(g_3=\tfrac34,f_3=\tfrac54,c_3=-\tfrac92)$:

   $\tfrac32 g+\tfrac52 f=c-\tfrac92.$

   From the first two, $c=4g+7$ and $f=4g+7$. Substituting into the third:

   $\tfrac32 g+\tfrac52(4g+7)=(4g+7)-\tfrac92\;\Rightarrow\;\tfrac{23}{2}g+\tfrac{35}{2}=4g+\tfrac52\;\Rightarrow\;g=-2.$

   Hence $c=-1,\ f=-1$, giving $S\equiv x^2+y^2-4x-2y-1=0$.

   The radical axis is $S-S_1=0$:

   $(x^2+y^2-4x-2y-1)-(x^2+y^2+4x-7)=-8x-2y+6=0,$

   $4x+y-3=0.$

   ✓Final answer

   (C) $4x + y - 3 = 0$.

4. TG EAPCET 2026Set eng-2026-05-10-AN1 markMCQ
   Q.Let $\vec{a} = \hat{i} + 2\hat{j} - 2\hat{k}$ and $\vec{b} = 6\hat{i} - 3\hat{j} + 2\hat{k}$ be two vectors. If $\vec{c}$ is a vector perpendicular to $\vec{a}$ and $\vec{c} \times \vec{b} = \hat{i} - 2\hat{j} - 6\hat{k}$, then the angle between the vectors $\vec{b}$ and $\vec{c}$ is (A) $\dfrac{\pi}{3}$ (B) $\cos^{-1} \left( \dfrac{29}{21\sqrt{2}} \right)$ (C) $\dfrac{\pi}{4}$ (D) $\cos^{-1} \left( \dfrac{23}{21\sqrt{2}} \right)$

   ›Reveal solutionSolution

   Solving the conditions gives $\vec c=4\hat i-\hat j+\hat k$, so $\cos\theta=\dfrac{\vec b\cdot\vec c}{|\vec b||\vec c|}=\dfrac{29}{21\sqrt2}$. Correct option: (B).

   Set up $\vec c=(x,y,z)$.

   Given $\vec a=\hat i+2\hat j-2\hat k$, $\vec b=6\hat i-3\hat j+2\hat k$.

   Perpendicularity to $\vec a$:

   $\vec c\cdot\vec a=x+2y-2z=0.\qquad(1)$

   Compute $\vec c\times\vec b$:

   $\vec c\times\vec b=(2y+3z,\;6z-2x,\;-3x-6y)=(1,\,-2,\,-6).$

   This gives:

   $2y+3z=1,\qquad 6z-2x=-2\;(\Rightarrow x=3z+1),\qquad -3x-6y=-6\;(\Rightarrow x+2y=2).$

   Solve. From $x+2y=2$ and (1), $x+2y-2z=0\Rightarrow 2-2z=0\Rightarrow z=1$.

   Then $x=3(1)+1=4$, and $2y+3(1)=1\Rightarrow y=-1$.

   $\vec c=4\hat i-\hat j+\hat k.$

   (Checks: $x+2y=2$ ✓, $\vec c\cdot\vec a=4-2-2=0$ ✓.)

   Angle between $\vec b$ and $\vec c$.

   $\vec b\cdot\vec c=6(4)+(-3)(-1)+2(1)=29,$

   $|\vec b|=\sqrt{36+9+4}=7,\qquad |\vec c|=\sqrt{16+1+1}=3\sqrt2.$

   $\cos\theta=\frac{29}{7\cdot 3\sqrt2}=\frac{29}{21\sqrt2}.$

   ✓Final answer

   $\theta=\cos^{-1}\!\left(\dfrac{29}{21\sqrt2}\right)$ — option (B).

5. TG EAPCET 2022Set eng-2022-07-20-AN1 markMCQ
   Q.The number of real values of $\alpha$ for which the pair of lines represented by $(\alpha^2 + 12|\alpha|)x^2 + 6xy + (18 - 21|\alpha|)y^2 = 0$ are at right angles to each other, is (A) 0 (B) 1 (C) 2 (D) 4

   ›Reveal solutionSolution

   For a pair of lines given by a homogeneous second-degree equation to be perpendicular, the condition is $a + b = 0$. Applying this to the given equation yields a quadratic in $|\alpha|$, giving two non-negative solutions, hence two real values of $\alpha$.

   The key idea is that a homogeneous equation of the form $ax^2 + 2hxy + by^2 = 0$ represents a pair of straight lines through the origin. For these lines to be perpendicular, the sum of the coefficients of $x^2$ and $y^2$ must be zero: $a + b = 0$. This is a standard result derived from the fact that if the slopes are $m_1$ and $m_2$, then $m_1 m_2 = -1$ for perpendicular lines, and from the relation $m_1 m_2 = a/b$ (with sign conventions), we get $a + b = 0$.

   Let’s apply this step by step.

   1. Identify the coefficients.

      The given equation is $(\alpha^2 + 12|\alpha|)x^2 + 6xy + (18 - 21|\alpha|)y^2 = 0$.

      Comparing with $ax^2 + 2hxy + by^2 = 0$, we have:

      $a = \alpha^2 + 12|\alpha|$,

      $2h = 6 \implies h = 3$,

      $b = 18 - 21|\alpha|$.

   2. Apply the perpendicular condition.

      For perpendicular lines, $a + b = 0$.

      So:

   $(\alpha^2 + 12|\alpha|) + (18 - 21|\alpha|) = 0$

   Simplify:

   $\alpha^2 + 12|\alpha| + 18 - 21|\alpha| = 0$

   $\alpha^2 - 9|\alpha| + 18 = 0$

   3. Solve for $|\alpha|$. Let $t = |\alpha| \geq 0$. The equation becomes:

   $t^2 - 9t + 18 = 0$

   Factor:

   $(t - 3)(t - 6) = 0$

   So $t = 3$ or $t = 6$. Both are non-negative, hence valid.

   4. Find the corresponding $\alpha$ values. $|\alpha| = 3$ gives $\alpha = 3$ or $\alpha = -3$. $|\alpha| = 6$ gives $\alpha = 6$ or $\alpha = -6$. That’s four distinct real values of $\alpha$ in total.

   Watch out

   A common mistake is to forget that $|\alpha|$ is non-negative and to treat the equation as a quadratic in $\alpha$ directly, which would give extraneous negative solutions. Always set $t = |\alpha|$ first.

   Tip

   Notice that the condition $a + b = 0$ is independent of $h$. So the $6xy$ term plays no role in the perpendicularity condition — it only ensures the lines are real and distinct (which we don’t need to check here since the question only asks for perpendicularity).

   ✓Final answer

   The number of real values of $\alpha$ is 4, which corresponds to option (D).

6. TG EAPCET 2021Set eng-2021-08-04-AN1 markMCQ
   Q.The polars of $(-1,\,2)$ with respect to the two circles $S_1 \equiv x^2 + y^2 + 6y + 7 = 0$ and $S_2 \equiv x^2 + y^2 + 6x + 1 = 0$ are (A) Parallel (B) coincident (C) Perpendicular (D) Intersecting at a non zero point

   ›Reveal solutionSolution

   The polar of a point with respect to a circle is a straight line whose equation is obtained by replacing $x^2 \to xx_1$, $y^2 \to yy_1$, $x \to \frac{x+x_1}{2}$, $y \to \frac{y+y_1}{2}$ in the circle's equation. For the given point $(-1,2)$ and the two circles, the polars turn out to be perpendicular lines.

   The concept of a polar is rooted in the geometry of tangents. If you draw the two tangents from an external point to a circle, the chord joining the points of contact is called the polar of that point. The equation of the polar of $(x_1, y_1)$ w.r.t. the circle $S \equiv x^2 + y^2 + 2gx + 2fy + c = 0$ is simply $T = 0$, i.e.:

   $xx_1 + yy_1 + g(x + x_1) + f(y + y_1) + c = 0$

   This is a straight line. So for each circle, we just plug in $(-1,2)$ into the $T=0$ form and get the two lines. Then we compare their slopes.

   ---

   1. Write the circles in standard form

      $S_1: x^2 + y^2 + 0x + 6y + 7 = 0$

      Here $2g = 0 \Rightarrow g = 0$, $2f = 6 \Rightarrow f = 3$, $c = 7$.

      $S_2: x^2 + y^2 + 6x + 0y + 1 = 0$

      Here $2g = 6 \Rightarrow g = 3$, $2f = 0 \Rightarrow f = 0$, $c = 1$.

   2. Write the polar of $(-1,2)$ for $S_1$

      Using $T = 0$:

   $x(-1) + y(2) + 0\cdot(x-1) + 3(y+2) + 7 = 0$

   Simplify:

   $-x + 2y + 3y + 6 + 7 = 0$

   $-x + 5y + 13 = 0$

   So the first polar is $L_1: -x + 5y + 13 = 0$, or $x - 5y - 13 = 0$. Its slope is $m_1 = \frac{1}{5}$.

   3. Write the polar of $(-1,2)$ for $S_2$

   $x(-1) + y(2) + 3(x-1) + 0\cdot(y+2) + 1 = 0$

   Simplify:

   $-x + 2y + 3x - 3 + 1 = 0$

   $2x + 2y - 2 = 0$

   Divide by 2: $x + y - 1 = 0$. So $L_2: x + y - 1 = 0$. Its slope is $m_2 = -1$.

   4. Check the relationship between the slopes $m_1 = \frac{1}{5}$, $m_2 = -1$. Product $m_1 \cdot m_2 = -\frac{1}{5} \neq -1$, so they are not perpendicular. They are clearly not parallel (slopes differ) and not coincident (different equations). Two non-parallel lines in a plane always intersect at exactly one point. That point is not the origin (check: $(0,0)$ satisfies neither), so they intersect at a non-zero point.

   Watch out

   A common mistake is to think that if $m_1 m_2 = -1$ then lines are perpendicular — but here the product is $-\frac{1}{5}$, not $-1$. Don't jump to "perpendicular" without checking.

   Tip

   You don't even need slopes: just check if the normals are proportional. For $L_1$ the normal is $(-1,5)$; for $L_2$ it's $(1,1)$. They are not scalar multiples (so not parallel/coincident), and their dot product $(-1)(1)+(5)(1)=4 \neq 0$ (so not perpendicular). Hence they must intersect at a unique point.

   ✓Final answer

   The polars intersect at a non-zero point, so the correct option is (D).

7. TG EAPCET 2023Set eng-2023-05-12-AN1 markMCQ
   Q.A line L has intercepts a and b on the coordinate axes. When the coordinate axes are rotated through an angle $\alpha$ keeping the origin fixed, the same line L has intercepts p and q on the new axes. Then (A) $a^2 + b^2 = p^2 + q^2$ (B) $a^2 + p^2 = b^2 + q^2$ (C) $\dfrac{1}{a^2} + \dfrac{1}{p^2} = \dfrac{1}{b^2} + \dfrac{1}{q^2}$ (D) $\dfrac{1}{a^2} + \dfrac{1}{b^2} = \dfrac{1}{p^2} + \dfrac{1}{q^2}$

   ›Reveal solutionSolution

   When coordinate axes are rotated, the geometric properties of a line that are independent of the coordinate system remain unchanged. The perpendicular distance from the origin to the line is one such invariant property, leading to the relation $\boxed{\dfrac{1}{a^2} + \dfrac{1}{b^2} = \dfrac{1}{p^2} + \dfrac{1}{q^2}}$.

   Concept and Intuition

   The problem describes a line L that is fixed in space. We are looking at this line from two different perspectives: first, with respect to an original set of coordinate axes, and then with respect to a new set of axes obtained by rotating the original ones around the origin. The key insight here is that the line itself does not move; only our reference frame changes.

   Any geometric property of the line that is independent of the choice of coordinate axes must remain the same. One such fundamental property is the perpendicular distance from the origin to the line. Since the origin remains fixed during the rotation, this distance must be invariant.

   Let's express the perpendicular distance from the origin to the line using its intercepts in both coordinate systems.

   Step-by-step Derivation

   1. Equation of the line in the original coordinate system: A line L with intercepts $a$ and $b$ on the $x$ and $y$ axes, respectively, has the equation in intercept form:

   $\frac{x}{a} + \frac{y}{b} = 1$

   To find the perpendicular distance from the origin $(0,0)$ to this line, we can rewrite the equation in the general form $Ax + By + C = 0$.
   

   $\frac{1}{a}x + \frac{1}{b}y - 1 = 0$

   > [!FORMULA]
   > The perpendicular distance $d$ from a point $(x_0, y_0)$ to a line $Ax + By + C = 0$ is given by:
   > $$ d = \frac{|Ax_0 + By_0 + C|}{\sqrt{A^2 + B^2}} $$
   For the origin $(0,0)$, the distance $d$ is:
   

   $d = \frac{|\frac{1}{a}(0) + \frac{1}{b}(0) - 1|}{\sqrt{\left(\frac{1}{a}\right)^2 + \left(\frac{1}{b}\right)^2}} = \frac{|-1|}{\sqrt{\frac{1}{a^2} + \frac{1}{b^2}}} = \frac{1}{\sqrt{\frac{1}{a^2} + \frac{1}{b^2}}}$

   Squaring both sides and taking the reciprocal, we get:
   

   $\frac{1}{d^2} = \frac{1}{a^2} + \frac{1}{b^2} \quad \text{(Equation 1)}$

   2. Equation of the line in the new coordinate system: When the coordinate axes are rotated through an angle $\alpha$ (keeping the origin fixed), the same line L has intercepts $p$ and $q$ on the new axes. Let the new coordinates be $(x', y')$. The equation of the line in the new system is:

   $\frac{x'}{p} + \frac{y'}{q} = 1$

   Similarly, the perpendicular distance from the origin $(0,0)$ to this line in the new coordinate system, let's call it $d'$, is:
   

   $d' = \frac{|\frac{1}{p}(0) + \frac{1}{q}(0) - 1|}{\sqrt{\left(\frac{1}{p}\right)^2 + \left(\frac{1}{q}\right)^2}} = \frac{|-1|}{\sqrt{\frac{1}{p^2} + \frac{1}{q^2}}} = \frac{1}{\sqrt{\frac{1}{p^2} + \frac{1}{q^2}}}$

   Squaring both sides and taking the reciprocal, we get:
   

   $\frac{1}{d'^2} = \frac{1}{p^2} + \frac{1}{q^2} \quad \text{(Equation 2)}$

   3. Equating the invariant distances: Since the line L is fixed in space and the origin is fixed, the perpendicular distance from the origin to the line must be the same regardless of the orientation of the coordinate axes. Therefore, $d = d'$. This implies $d^2 = d'^2$, and consequently $\frac{1}{d^2} = \frac{1}{d'^2}$. Substituting the expressions from Equation 1 and Equation 2:

   $\frac{1}{a^2} + \frac{1}{b^2} = \frac{1}{p^2} + \frac{1}{q^2}$

   This relation holds true for any rotation of the coordinate axes about the origin.
   

   Tip

   This problem can also be solved using the coordinate transformation formulas for rotation. If $(x,y)$ are the original coordinates and $(x',y')$ are the new coordinates after rotating the axes by an angle $\alpha$, then $x = x' \cos \alpha - y' \sin \alpha$ and $y = x' \sin \alpha + y' \cos \alpha$. Substituting these into the original line equation $\frac{x}{a} + \frac{y}{b} = 1$ and rearranging to the form $\frac{x'}{p} + \frac{y'}{q} = 1$ would also yield the same result, but the invariant property method is more direct and conceptually elegant.

   ---

   ✓Final answer

   The correct relation between the intercepts is $\boxed{\dfrac{1}{a^2} + \dfrac{1}{b^2} = \dfrac{1}{p^2} + \dfrac{1}{q^2}}$, which corresponds to option (D).

8. TG EAPCET 2025Set eng-2025-05-03-AN1 markMCQ
   Q.The number of values of ‘k’ for which the points $(-4,9,k)$, $(-1,6,k)$, $(0,7,10)$ form a right-angled isosceles triangle is (A) 0 (B) 1 (C) 2 (D) 4

   ›Reveal solutionSolution

   The key idea is to use the distance formula and the conditions for a right-angled isosceles triangle (two equal sides and a right angle) to set up equations in $k$. Solving these yields exactly two valid values, so the correct option is (C).

   We are given three points in 3D space:

   $A(-4,9,k)$, $B(-1,6,k)$, $C(0,7,10)$.

   We need the number of values of $k$ for which triangle $ABC$ is right-angled and isosceles.

   Concept and Intuition

   A triangle is right-angled isosceles if it has two equal sides and the angle between them is $90^\circ$. In coordinate geometry, we can check this using distances:

   • Two sides must be equal in length.
   • The Pythagorean theorem must hold for those two sides and the third side (the hypotenuse). Since the points have a variable $k$ in the $z$-coordinate for $A$ and $B$, the distances will depend on $k$. We compute all three squared distances, set up conditions, and solve for $k$.

   Step-by-step solution

   1. Compute squared distances Let $d_{AB}^2$, $d_{BC}^2$, $d_{AC}^2$ be the squared distances between the points.

   $d_{AB}^2 = (-1+4)^2 + (6-9)^2 + (k-k)^2 = 3^2 + (-3)^2 + 0^2 = 9+9 = 18.$

   Notice $d_{AB}^2$ is constant (independent of $k$).

   $d_{BC}^2 = (0+1)^2 + (7-6)^2 + (10-k)^2 = 1^2 + 1^2 + (10-k)^2 = 2 + (10-k)^2.$

   $d_{AC}^2 = (0+4)^2 + (7-9)^2 + (10-k)^2 = 4^2 + (-2)^2 + (10-k)^2 = 16+4 + (10-k)^2 = 20 + (10-k)^2.$

   2. Identify possible equal sides

      Since $d_{AB}^2 = 18$ is constant, the equal sides could be:

      • Case I: $AB = BC$
      • Case II: $AB = AC$
      • Case III: $BC = AC$

      We must also enforce the right-angle condition: the square of the longest side equals the sum of squares of the other two.

   3. Case I: $AB = BC$

   $18 = 2 + (10-k)^2 \quad\Rightarrow\quad (10-k)^2 = 16 \quad\Rightarrow\quad 10-k = \pm 4.$

   So $k = 6$ or $k = 14$.

   Now check the right-angle condition. If $AB = BC$, the right angle could be at $B$ (between $AB$ and $BC$) or at $A$ or $C$. But the equal sides meet at the vertex where the right angle is (in an isosceles right triangle, the equal sides are the legs). So the right angle should be at the vertex where the two equal sides meet.

   • If $AB = BC$, the equal sides meet at $B$. So we need $AB^2 + BC^2 = AC^2$. For $k=6$: $AB^2 = 18$, $BC^2 = 2 + (4)^2 = 2+16=18$, $AC^2 = 20 + (4)^2 = 20+16=36$. Check: $18+18=36$ — works. For $k=14$: $BC^2 = 2 + (-4)^2 = 18$, $AC^2 = 20 + (-4)^2 = 36$. $18+18=36$ — works. So both $k=6$ and $k=14$ satisfy Case I.

   4. Case II: $AB = AC$

   $18 = 20 + (10-k)^2 \quad\Rightarrow\quad (10-k)^2 = -2,$

   impossible (square cannot be negative). So no solutions here.

   5. Case III: $BC = AC$

   $2 + (10-k)^2 = 20 + (10-k)^2 \quad\Rightarrow\quad 2 = 20,$

   impossible. So no solutions here either.

   6. Conclusion Only Case I yields solutions, and we found two distinct values: $k=6$ and $k=14$. Both produce a right-angled isosceles triangle.

   Watch out

   A common mistake is to forget that the equal sides must be the legs of the right triangle. If we had only set two sides equal without checking the right-angle condition at the correct vertex, we might incorrectly accept extra values or miss verifying.

   Tip

   Notice that $AB$ is constant, which simplifies the problem: only one case (where $AB$ is one of the equal legs) is possible. The other cases fail immediately because the constant term mismatch gives no real solutions.

   ✓Final answer

   The correct option is (C).

   ANSWER: C

9. TG EAPCET 2022Set eng-2022-07-18-AN1 markMCQ
   Q.If $\theta$ is the angle between the circles $x^2 + y^2 - 2x - 4y - 4 = 0$ and $x^2 + y^2 - 8x - 12y + 43 = 0$ then $|7\sec\theta - 18\cos\theta| =$ (A) 11 (B) 9 (C) 0 (D) 1

   ›Reveal solutionSolution

   The angle between two circles is found using their radii and the distance between their centers. We calculate $\cos\theta = -7/18$ and then substitute this into the given expression, which simplifies to $\boxed{11}$.

   The angle between two intersecting circles is defined as the angle between their tangents at a point of intersection. This angle can be determined using the properties of the circles: their centers and radii.

   Consider the two circles intersecting at a point $P$. Let $C_1$ and $C_2$ be their centers, and $r_1$ and $r_2$ be their respective radii. The lines $C_1P$ and $C_2P$ are radii. The tangent to the first circle at $P$ is perpendicular to $C_1P$, and the tangent to the second circle at $P$ is perpendicular to $C_2P$. The angle $\theta$ between the tangents is equal to the angle between the radii $C_1P$ and $C_2P$, or its supplement.

   Let $d$ be the distance between the centers $C_1$ and $C_2$. In the triangle $\triangle C_1PC_2$, the sides are $r_1$, $r_2$, and $d$. By the cosine rule, if $\theta$ is the angle $\angle C_1PC_2$, then:

   $d^2 = r_1^2 + r_2^2 - 2r_1r_2 \cos\theta$

   Rearranging this gives the formula for $\cos\theta$:

   The cosine of the angle $\theta$ between two circles with centers $C_1, C_2$ and radii $r_1, r_2$, where $d$ is the distance between their centers, is given by:

   $\cos\theta = \frac{r_1^2 + r_2^2 - d^2}{2r_1r_2}$

   This formula can also be expressed in terms of the general equation of a circle $x^2 + y^2 + 2gx + 2fy + c = 0$. For two circles $x^2 + y^2 + 2g_1x + 2f_1y + c_1 = 0$ and $x^2 + y^2 + 2g_2x + 2f_2y + c_2 = 0$:

   [!FORMULA]

   $\cos\theta = \frac{2g_1g_2 + 2f_1f_2 - c_1 - c_2}{2r_1r_2}$

   We will use this formula to find $\cos\theta$.

   Here is the step-by-step solution:

   1. Identify the centers and radii of the circles.

      The general equation of a circle is $x^2 + y^2 + 2gx + 2fy + c = 0$, with center $(-g, -f)$ and radius $r = \sqrt{g^2 + f^2 - c}$.

      For the first circle, $x^2 + y^2 - 2x - 4y - 4 = 0$:

      $2g_1 = -2 \implies g_1 = -1$

      $2f_1 = -4 \implies f_1 = -2$

      $c_1 = -4$

      Center $C_1 = (-g_1, -f_1) = (1, 2)$

      Radius $r_1 = \sqrt{(-1)^2 + (-2)^2 - (-4)} = \sqrt{1 + 4 + 4} = \sqrt{9} = 3$.

      For the second circle, $x^2 + y^2 - 8x - 12y + 43 = 0$:

      $2g_2 = -8 \implies g_2 = -4$

      $2f_2 = -12 \implies f_2 = -6$

      $c_2 = 43$

      Center $C_2 = (-g_2, -f_2) = (4, 6)$

      Radius $r_2 = \sqrt{(-4)^2 + (-6)^2 - 43} = \sqrt{16 + 36 - 43} = \sqrt{52 - 43} = \sqrt{9} = 3$.

   2. Calculate the cosine of the angle $\theta$.

      Using the formula $\cos\theta = \frac{2g_1g_2 + 2f_1f_2 - c_1 - c_2}{2r_1r_2}$:

   $\cos\theta = \frac{2(-1)(-4) + 2(-2)(-6) - (-4) - 43}{2(3)(3)}$

   $\cos\theta = \frac{8 + 24 + 4 - 43}{18}$

   $\cos\theta = \frac{36 - 43}{18}$

   $\cos\theta = \frac{-7}{18}$

   3. Calculate the required expression $|7\sec\theta - 18\cos\theta|$.

      We have $\cos\theta = -\frac{7}{18}$.

      Therefore, $\sec\theta = \frac{1}{\cos\theta} = \frac{1}{-7/18} = -\frac{18}{7}$.

      Now substitute these values into the expression:

   $|7\sec\theta - 18\cos\theta| = \left|7\left(-\frac{18}{7}\right) - 18\left(-\frac{7}{18}\right)\right|$

   $= |-18 - (-7)|$

   $= |-18 + 7|$

   $= |-11|$

   $= 11$

   ✓Final answer

   The value of $|7\sec\theta - 18\cos\theta|$ is $\boxed{11}$.

10. TG EAPCET 2025Set eng-2025-05-03-AN1 markMCQ
    Q.If $(\alpha, \beta)$ is the centre of the circle which passes through the point $(1, -1)$ and cuts the circles $x^2 + y^2 + 2x - 3y - 5 = 0$, $x^2 + y^2 - 3x + 2y + 1 = 0$ orthogonally, then $\alpha - 5\beta =$ (A) $-10$ (B) $5$ (C) $-11$ (D) $10$

    ›Reveal solutionSolution

    Centre $(\alpha,\beta)=\left(-1,\,-\tfrac{11}{5}\right)$, so $\alpha-5\beta=10$.

    Let the required circle be $x^2+y^2+2gx+2fy+c=0$, with centre $(\alpha,\beta)=(-g,-f)$.

    Orthogonality with $x^2+y^2+2x-3y-5=0$ (where $g_1=1,\,f_1=-\tfrac32,\,c_1=-5$):

    $2g(1)+2f\!\left(-\tfrac32\right)=c-5\;\Rightarrow\;2g-3f=c-5.$

    Orthogonality with $x^2+y^2-3x+2y+1=0$ (where $g_2=-\tfrac32,\,f_2=1,\,c_2=1$):

    $-3g+2f=c+1.$

    Passing through $(1,-1)$: $\,2+2g-2f+c=0\Rightarrow c=-2-2g+2f.$

    Substituting $c$ into the two orthogonality relations gives

    $4g-5f=-7,\qquad -g=-1\Rightarrow g=1,$

    and then $f=\tfrac{11}{5}$. Thus $\alpha=-g=-1$ and $\beta=-f=-\tfrac{11}{5}$.

    $\alpha-5\beta=-1-5\!\left(-\tfrac{11}{5}\right)=-1+11=10.$

    ✓Final answer

    ANSWER: D — $\;\alpha-5\beta=10$.

11. TG EAPCET 2026Set eng-2026-05-10-FN1 markMCQ
    Q.If the circles $x^2 + y^2 - 10x + 8y + 5 = 0$ and $x^2 + y^2 + 6x - 4y + c = 0$ cut each other orthogonally, then the sum of radii of these circles is (A) 14 (B) 10 (C) 12 (D) 9

    ›Reveal solutionSolution

    Two circles cut orthogonally if the square of the distance between their centers equals the sum of the squares of their radii. This leads to the condition $2g_1g_2 + 2f_1f_2 = c_1 + c_2$. Applying this, we find the value of $c$, then calculate the radii of both circles and sum them to get $\boxed{14}$.

    Concept and Intuition

    When two circles intersect, they are said to cut each other orthogonally if their tangents at any point of intersection are perpendicular. This geometric condition has a powerful algebraic equivalent that simplifies calculations.

    Consider two circles with centers $C_1$ and $C_2$ and radii $r_1$ and $r_2$ respectively. Let $P$ be a point of intersection.

    The radius $C_1P$ is perpendicular to the tangent of the first circle at $P$.

    Similarly, the radius $C_2P$ is perpendicular to the tangent of the second circle at $P$.

    Since the tangents themselves are perpendicular, it follows that the radii $C_1P$ and $C_2P$ must also be perpendicular.

    Therefore, the triangle $C_1PC_2$ is a right-angled triangle with the right angle at $P$.

    By the Pythagorean theorem, the square of the distance between the centers, $(C_1C_2)^2$, must be equal to the sum of the squares of the radii, $r_1^2 + r_2^2$.

    For two circles $S_1: x^2 + y^2 + 2g_1x + 2f_1y + c_1 = 0$ and $S_2: x^2 + y^2 + 2g_2x + 2f_2y + c_2 = 0$ to cut orthogonally, the condition is:

    $2g_1g_2 + 2f_1f_2 = c_1 + c_2$

    ›Proof

    Let the centers of the circles be $C_1 = (-g_1, -f_1)$ and $C_2 = (-g_2, -f_2)$.

    The radii are $r_1 = \sqrt{g_1^2 + f_1^2 - c_1}$ and $r_2 = \sqrt{g_2^2 + f_2^2 - c_2}$.

    The condition for orthogonal intersection is $(C_1C_2)^2 = r_1^2 + r_2^2$.

    $(C_1C_2)^2 = (-g_1 - (-g_2))^2 + (-f_1 - (-f_2))^2$

    $= (g_2 - g_1)^2 + (f_2 - f_1)^2$

    $= g_2^2 - 2g_1g_2 + g_1^2 + f_2^2 - 2f_1f_2 + f_1^2$

    And $r_1^2 + r_2^2 = (g_1^2 + f_1^2 - c_1) + (g_2^2 + f_2^2 - c_2)$.

    Equating these:

    $g_1^2 + g_2^2 - 2g_1g_2 + f_1^2 + f_2^2 - 2f_1f_2 = g_1^2 + f_1^2 - c_1 + g_2^2 + f_2^2 - c_2$

    Cancelling $g_1^2, g_2^2, f_1^2, f_2^2$ from both sides:

    $-2g_1g_2 - 2f_1f_2 = -c_1 - c_2$

    Multiplying by $-1$:

    $2g_1g_2 + 2f_1f_2 = c_1 + c_2$

    Step-by-step Derivations

    1. Identify parameters for the first circle:

       The first circle is given by $x^2 + y^2 - 10x + 8y + 5 = 0$.

       Comparing this with the general equation $x^2 + y^2 + 2g_1x + 2f_1y + c_1 = 0$:

       $2g_1 = -10 \implies g_1 = -5$

       $2f_1 = 8 \implies f_1 = 4$

       $c_1 = 5$

       The radius of the first circle, $r_1$, is given by $\sqrt{g_1^2 + f_1^2 - c_1}$.

       $r_1 = \sqrt{(-5)^2 + (4)^2 - 5} = \sqrt{25 + 16 - 5} = \sqrt{41 - 5} = \sqrt{36} = 6$.

    2. Identify parameters for the second circle:

       The second circle is given by $x^2 + y^2 + 6x - 4y + c = 0$.

       Comparing this with $x^2 + y^2 + 2g_2x + 2f_2y + c_2 = 0$:

       $2g_2 = 6 \implies g_2 = 3$

       $2f_2 = -4 \implies f_2 = -2$

       $c_2 = c$

       The radius of the second circle, $r_2$, is $\sqrt{g_2^2 + f_2^2 - c_2}$.

       $r_2 = \sqrt{(3)^2 + (-2)^2 - c} = \sqrt{9 + 4 - c} = \sqrt{13 - c}$.

    3. Apply the condition for orthogonal intersection to find $c$:

       Since the circles cut each other orthogonally, we use the condition $2g_1g_2 + 2f_1f_2 = c_1 + c_2$.

       Substitute the values we found:

       $2(-5)(3) + 2(4)(-2) = 5 + c$

       $-30 - 16 = 5 + c$

       $-46 = 5 + c$

       $c = -46 - 5$

       $c = -51$

    4. Calculate the radius of the second circle using the value of $c$:

       Now that we have $c = -51$, we can find $r_2$:

       $r_2 = \sqrt{13 - c} = \sqrt{13 - (-51)} = \sqrt{13 + 51} = \sqrt{64} = 8$.

    5. Calculate the sum of the radii:

       The sum of the radii is $r_1 + r_2$.

       Sum $= 6 + 8 = 14$.

    ✓Final answer

    The sum of the radii of these circles is $\boxed{14}$.

12. TG EAPCET 2023Set eng-2023-05-13-AN1 markMCQ
    Q.Let $A(1,3)$ and $B(2,5)$ be two points and $C(h, k)$ be a point such that $BC$ is perpendicular to $AC$. If $\angle CAB = \angle CBA$, then $h =$ (A) $\dfrac{24}{5}$ or $\dfrac{7}{2}$ (B) $\dfrac{2}{5}$ or $\dfrac{7}{2}$ (C) $\dfrac{1}{2}$ or $\dfrac{5}{2}$ (D) $\dfrac{24}{5}$ or $\dfrac{2}{5}$

    ›Reveal solutionSolution

    $\angle CAB=\angle CBA$ with $\angle ACB=90^\circ$ makes $\triangle ACB$ right-isosceles; $C$ is the apex, giving $h=\tfrac12$ or $\tfrac52$.

    Let $A(1,3)$, $B(2,5)$, $C(h,k)$.

    Isosceles condition: $\angle CAB=\angle CBA$ means the sides opposite these equal angles are equal, i.e. $CB=CA$, so $C$ lies on the perpendicular bisector of $AB$.

    Right angle: $BC\perp AC$ gives $\angle ACB=90^\circ$. With $CA=CB$, $\triangle ACB$ is right-angled and isosceles at $C$, so $C$ is the apex whose distance from the midpoint $M$ of $AB$ equals $\tfrac12|AB|$ (the median to the hypotenuse).

    Geometry:

    $M=\left(\tfrac{3}{2},4\right),\qquad |AB|=\sqrt{(2-1)^2+(5-3)^2}=\sqrt5,\qquad CM=\tfrac{\sqrt5}{2}.$

    A unit vector perpendicular to $AB=(1,2)$ is $\dfrac{(2,-1)}{\sqrt5}$. Hence

    $C=M\pm\frac{\sqrt5}{2}\cdot\frac{(2,-1)}{\sqrt5}=\left(\tfrac{3}{2},4\right)\pm\left(1,-\tfrac12\right).$

    This gives $C=\left(\tfrac52,\tfrac72\right)$ or $C=\left(\tfrac12,\tfrac92\right)$, so

    $h=\frac52\quad\text{or}\quad h=\frac12.$

    ✓Final answer

    $h=\dfrac{1}{2}$ or $\dfrac{5}{2}$ — option (C).