Q.If $l_1, m_1, n_1$; $l_2, m_2, n_2$; $l_3, m_3, n_3$ are the direction cosines of three mutually perpendicular lines, prove that the line whose direction cosines are proportional to $l_1 + l_2 + l_3$, $m_1 + m_2 + m_3$, $n_1 + n_2 + n_3$ makes equal angles with them.

Showing the 12 most recent of 21 on this concept.

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

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2. 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: …

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3. 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: $$$$

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4. 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 $h = 6$. …

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5. 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 $2y+3(1)=1\Rightarrow y=-1$.

   $\vec c=4\hat i-\hat j+\hat k.$ …

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6. TG EAPCET 2022Set eng-2022-07-18-AN1 markMCQ
   Q.Let $S = \dfrac{x^2}{a^2} + \dfrac{y^2}{b^2} - 1 = 0$, $S' = \dfrac{x^2}{\alpha^2} + \dfrac{y^2}{\beta^2} - 1 = 0$ be two intersecting ellipses. If $P(\alpha \cos \theta, b \sin \theta)$ and $Q\left(\alpha \cos \left(\dfrac{\pi}{2} + \theta\right), b \sin \left(\dfrac{\pi}{2} + \theta\right)\right)$ are their points of intersection then $\dfrac{1}{2}(a^2 \beta^2 + b^2 \alpha^2) =$ (A) $a^2 b^2$ (B) $\alpha^2 + \beta^2$ (C) $a^2 + b^2$ (D) $\alpha^2 \beta^2$

   ›Reveal solutionSolution

   Both intersection points lie on S'; adding their equations gives ½(a²β²+b²α²)=α²β².

   Take the intersection points parameterised on the first ellipse and impose that they satisfy S': x²/α² + y²/β² = 1. Writing the condition for P at parameter θ and for Q at θ+π/2 and adding, the sin²+cos² terms combine to give a² …

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7. 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 $\alpha = -3$. …

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8. 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. …

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9. 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 $k=14$: …

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10. TG EAPCET 2022Set eng-2022-07-18-FN1 markMCQ
    Q.In triangle ABC, if BC is the hypotenuse, then $r_2 + r_3 =$ (A) $r_1 + r$ (B) $a$ (C) $r - r_1$ (D) $2(R + r)$

    ›Reveal solutionSolution

    In a right triangle with BC as the hypotenuse, the exradii opposite the legs sum to the hypotenuse itself, so $r_2 + r_3 = a$.

    Concept & Intuition

    We have triangle $ABC$ with $BC$ as the hypotenuse, so $\angle A = 90^\circ$. The notation $r_1, r_2, r_3$ typically denotes the exradii opposite vertices $A, B, C$ respectively, and $r$ is the inradius. The problem asks for $r_2 + r_3$ — the sum of the exradii opposite the legs. In a right triangle, geometry often simplifies because the hypotenuse is the diameter of the circumcircle, and the inradius has a neat expression. The key is to use standard formulas for exradii in terms of sides and area, then substitute the right-angle condition.

    Step-by-step reasoning

    1. Set up notation

       Let sides be: $BC = a$ (hypotenuse), $CA = b$, $AB = c$. Since $\angle A = 90^\circ$, we have $a^2 = b^2 + c^2$.

       Semi-perimeter: $s = \frac{a+b+c}{2}$.

       Area: $\Delta = \frac{1}{2}bc$.

    2. Recall exradii formulas

       The exradius opposite vertex $A$ (touching side $BC$) is $r_1 = \frac{\Delta}{s-a}$.

       Opposite $B$: $r_2 = \frac{\Delta}{s-b}$.

       Opposite $C$: $r_3 = \frac{\Delta}{s-c}$.

       Inradius: $r = \frac{\Delta}{s}$.

    3. Compute $r_2 + r_3$

    $r_2 + r_3 = \frac{\Delta}{s-b} + \frac{\Delta}{s-c} = \Delta \cdot \frac{(s-c)+(s-b)}{(s-b)(s-c)} = \Delta \cdot \frac{2s - (b+c)}{(s-b)(s-c)}.$

    But $2s = a+b+c$, so $2s - (b+c) = a$. Hence:

    $r_2 + r_3 = \Delta \cdot \frac{a}{(s-b)(s-c)}.$

    4. Simplify $(s-b)(s-c)$

    $s-b = \frac{a+b+c}{2} - b = \frac{a - b + c}{2},$

    $s-c = \frac{a+b+c}{2} - c = \frac{a + b - c}{2}.$

    Multiply:

    $(s-b)(s-c) = \frac{(a - b + c)(a + b - c)}{4}.$

    Notice $(a - b + c)(a + b - c) = a^2 - (b-c)^2 = a^2 - (b^2 - 2bc + c^2) = (a^2 - b^2 - c^2) + 2bc$.

    Since $a^2 = b^2 + c^2$, this becomes $0 + 2bc = 2bc$.

    Therefore:

    $$$$

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11. 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:

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

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12. 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, $x^2 + y^2 - 8x - 12y + 43 = 0$:

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

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