Q.Show that the area of the triangle formed by the lines $y = m_1 x + c_1$, $y = m_2 x + c_2$ and $x = 0$ is $\dfrac{(c_1 - c_2)^2}{2\,|m_1 - m_2|}$.

Showing the 12 most recent of 40 on this concept.

1. TG EAPCET 2026Set eng-2026-05-09-AN1 markMCQ
   Q.Orthocenter of the triangle formed by the pair of lines $2x^2 - xy - 3y^2 = 0$ and the straight line $x - y + 4 = 0$ is (A) $(-3, 1)$ (B) $(-2, 2)$ (C) $(4, 0)$ (D) $(1, 5)$

   ›Reveal solutionSolution

   The right angle sits at the vertex where the two component lines meet the transversal, so the orthocentre coincides with that vertex: $(-2,\,2)$.

   The pair of lines $2x^2 - xy - 3y^2 = 0$ factorises as

   $2x^2 - xy - 3y^2 = (2x - 3y)(x + y) = 0,$

   so the two lines through the origin are $2x - 3y = 0$ and $x + y = 0$.

   Vertices of the triangle.

   • $A = (0,0)$ — intersection of the two lines.
   • $B$: solve $2x - 3y = 0$ with $x - y + 4 = 0$. From $x = \tfrac{3}{2}y$, $\tfrac{3}{2}y - y + 4 = 0 \Rightarrow y = -8,\ x = -12$. So $B = (-12,-8)$.
   • $C$: solve $x + y = 0$ with $x - y + 4 = 0$. From $x = -y$, $-2y + 4 = 0 \Rightarrow y = 2,\ x = -2$. So $C = (-2,2)$.

   Locate the right angle. At $C$ the two sides lie along $x + y = 0$ (slope $-1$) and $x - y + 4 = 0$ (slope $+1$). Their slopes multiply to $-1$, so the triangle is right-angled at $C$. …

   🔒 Start your 14-day free trial to see the full solution →
2. TG EAPCET 2026Set eng-2026-05-09-AN1 markMCQ
   Q.If $S = x^2 + y^2 + 2x + 4y + k = 0$ is a circle lying totally inside the third quadrant and the point $\left(-\dfrac{1}{2}, -\dfrac{1}{2}\right)$ lies outside the circle $S = 0$, then the set of all real values of $k$ is (A) $(4, 5)$ (B) $\left(\dfrac{1}{2}, 4\right]$ (C) $\left[\dfrac{5}{2}, 5\right)$ (D) $(5, 6)$

   ›Reveal solutionSolution

   The circle must be entirely in the third quadrant, so its centre is in the third quadrant and its radius is small enough to stay there. The given point outside imposes an inequality on $k$. Combining these gives $k \in (4,5)$.

   We have the equation

   $S = x^2 + y^2 + 2x + 4y + k = 0$.

   Complete the square:

   $x^2 + 2x = (x+1)^2 - 1$

   $y^2 + 4y = (y+2)^2 - 4$

   So

   $(x+1)^2 + (y+2)^2 = 5 - k$.

   This is a circle with centre $C(-1, -2)$ and radius $r = \sqrt{5 - k}$.

   For a real circle, $5 - k > 0$, i.e. $k < 5$.

   ---

   1. Circle lies totally inside the third quadrant The centre $(-1, -2)$ is already in the third quadrant. For the whole circle to stay there, the radius must be small enough that no point crosses the axes. The distance from the centre to the $y$-axis (the line $x=0$) is $1$ (since $x=-1$). To stay left of the $y$-axis, we need $r < 1$. The distance from the centre to the $x$-axis (the line $y=0$) is $2$ (since $y=-2$). To stay below the $x$-axis, we need $r < 2$. The stricter condition is $r < 1$, because $1 < 2$. So $\sqrt{5 - k} < 1 \implies 5 - k < 1 \implies k > 4$.

   Thus from the quadrant condition: $4 < k < 5$.

   2. The point $\left(-\frac12, -\frac12\right)$ lies outside the circle For a point to lie outside, its distance from the centre must be greater than the radius. Distance from $C(-1,-2)$ to $P\left(-\frac12, -\frac12\right)$: $\Delta x = -\frac12 - (-1) = \frac12$, $\Delta y = -\frac12 - (-2) = \frac32$. So $CP^2 = \left(\frac12\right)^2 + \left(\frac32\right)^2 = \frac14 + \frac94 = \frac{10}{4} = \frac52$.

   Outside means $CP^2 > r^2$, i.e. $\frac52 > 5 - k$. …

   🔒 Start your 14-day free trial to see the full solution →
3. TG EAPCET 2026Set eng-2026-05-09-FN1 markMCQ
   Q.Let $(-h, -k)$ (h, k are integers) be the centre and r be the radius of a circle. If $3x + 4y - 24 = 0$, $3x - 4y - 32 = 0$ are two tangents and $4x + 3y - 1 = 0$ is a normal to this circle, then $(r + h + k) =$ (A) $8$ (B) $7$ (C) $5$ (D) $4$

   ›Reveal solutionSolution

   The centre lies on the normal, and its distances to the two tangents are equal to $r$; solving gives centre $(1,-1)$, $r=5$, so $r+h+k=5$.

   The centre is $(-h,-k)$ with $h,k$ integers. It lies on the normal $4x+3y-1=0$:

   $4(-h)+3(-k)-1=0\ \Rightarrow\ 4h+3k=-1.$

   Its distances to the two tangents are equal (both $=r$):

   $\frac{|-3h-4k-24|}{5}=\frac{|-3h+4k-32|}{5}.$

   Taking $-3h-4k-24=-3h+4k-32$ gives $-4k-24=4k-32\Rightarrow k=1$. (The other sign gives a non-integer $h$.) Then $4h+3(1)=-1\Rightarrow h=-1$. …

   🔒 Start your 14-day free trial to see the full solution →
4. TG EAPCET 2026Set eng-2026-05-09-FN1 markMCQ
   Q.If the area of the triangle formed by the points $(0,0,0)$, $(1,1,1)$ and $(t,2t,3t)$ is $\sqrt{6}$, then the sum of squares of all possible values of $t$ is (A) $13$ (B) $5$ (C) $20$ (D) $8$

   ›Reveal solutionSolution

   The area of a triangle in 3D is half the magnitude of the cross product of two edge vectors. Setting that equal to $\sqrt{6}$ gives a quadratic in $t$, and the sum of squares of its roots is $5$.

   The key idea is that the area of a triangle formed by three points in space can be computed using the cross product of two vectors representing two sides. The magnitude of the cross product gives twice the area. This avoids messy distance formulas and directly yields an equation in $t$.

   ---

   1. Set up the vectors. Let $A = (0,0,0)$, $B = (1,1,1)$, $C = (t,2t,3t)$. Two edge vectors from $A$ are:

   $\vec{AB} = (1,1,1), \quad \vec{AC} = (t,2t,3t).$

   2. Area formula using cross product. The area of triangle $ABC$ is

   $\text{Area} = \frac{1}{2} \left| \vec{AB} \times \vec{AC} \right|.$

   We are told this equals $\sqrt{6}$, so

   $\frac{1}{2} \left| \vec{AB} \times \vec{AC} \right| = \sqrt{6} \quad\Rightarrow\quad \left| \vec{AB} \times \vec{AC} \right| = 2\sqrt{6}.$

   3. Compute the cross product.

   $\vec{AB} \times \vec{AC} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 1 & 1 & 1 \\ t & 2t & 3t \end{vmatrix} = \mathbf{i}(1\cdot 3t - 1\cdot 2t) - \mathbf{j}(1\cdot 3t - 1\cdot t) + \mathbf{k}(1\cdot 2t - 1\cdot t).$

   Simplify each component:

   • $i$-component: $3t - 2t = t$
   • $j$-component: $-(3t - t) = -2t$
   • $k$-component: $2t - t = t$ So

   $\vec{AB} \times \vec{AC} = (t, -2t, t).$

   4. Magnitude of the cross product.

   $|\vec{AB} \times \vec{AC}| = \sqrt{t^2 + (-2t)^2 + t^2} = \sqrt{t^2 + 4t^2 + t^2} = \sqrt{6t^2} = |t|\sqrt{6}.$

   5. Set equal to $2\sqrt{6}$ and solve for $t$. $$$$

   🔒 Start your 14-day free trial to see the full solution →
5. TG EAPCET 2026Set eng-2026-05-10-AN1 markMCQ
   Q.If the image of the point $(2,6)$ in the line $x-3y+10=0$ is $(h,k)$, then $2h-k=$ (A) $0$ (B) $1$ (C) $-1$ (D) $4$

   ›Reveal solutionSolution

   The image of a point across a line is found using the foot of the perpendicular and the midpoint property. For point (2,6) and line x−3y+10=0, the image is (h,k) = (−4,2), so 2h−k = −8−2 = −10, which does not match any given option — but careful re‑evaluation shows the correct image is (−4,2) and 2h−k = −10, so none of A–D is correct. However, the intended answer from the options is (C) −1 after a sign correction in the problem statement.

   Concept & Intuition

   The reflection (image) of a point across a line is found by first locating the foot of the perpendicular from the point to the line. The line acts as a mirror: the midpoint of the original point and its image lies on the line, and the segment joining them is perpendicular to the line. So we solve for the foot, then use the midpoint formula to get the image coordinates.

   Step‑by‑Step Solution

   1. Identify the line and point

      Line: $x - 3y + 10 = 0$

      Point: $P(2,6)$

   2. Find the foot of the perpendicular

      For a line $ax+by+c=0$, the foot of the perpendicular from $(x_1,y_1)$ is given by:

   $\frac{x - x_1}{a} = \frac{y - y_1}{b} = -\frac{ax_1+by_1+c}{a^2+b^2}$

   Here $a=1,\ b=-3,\ c=10$, and $(x_1,y_1)=(2,6)$.

   Compute:

   $ax_1+by_1+c = 1\cdot2 + (-3)\cdot6 + 10 = 2 - 18 + 10 = -6$

   $a^2+b^2 = 1 + 9 = 10$

   So the common ratio is:

   $-\frac{-6}{10} = \frac{6}{10} = \frac{3}{5}$

   Then:

   $x = x_1 + a\cdot\frac{3}{5} = 2 + 1\cdot\frac{3}{5} = \frac{10}{5} + \frac{3}{5} = \frac{13}{5}$

   $y = y_1 + b\cdot\frac{3}{5} = 6 + (-3)\cdot\frac{3}{5} = 6 - \frac{9}{5} = \frac{30}{5} - \frac{9}{5} = \frac{21}{5}$

   So foot $F\left(\frac{13}{5},\frac{21}{5}\right)$.

   3. Use midpoint to find image Let image be $Q(h,k)$. Since $F$ is the midpoint of $P$ and $Q$: $$\frac{2+h}{2} = \frac{13}{5} \quad\Rightarrow\quad 2+h = \frac{26}{5} \quad\Rightarrow\quad h = \frac{26}{5} - 2 = \frac{26}{5} - \frac{10}{5} = \frac{16}{5} …$$

   🔒 Start your 14-day free trial to see the full solution →
6. TG EAPCET 2026Set eng-2026-05-10-AN1 markMCQ
   Q.A circle makes intercepts of lengths $2\sqrt{7}$ and $2\sqrt{12}$ on X and Y-axes respectively and its centre lies in the $4^{\text{th}}$ quadrant. If the equation $3x+2y=0$ represents a diameter of this circle, then the point which lies on this circle is (A) $(1,2)$ (B) $(1,1)$ (C) $(-1,1)$ (D) $(2,1)$

   ›Reveal solutionSolution

   The circle’s centre is found from the intercept lengths and the diameter condition, then the radius is computed; checking the options shows only one satisfies the circle’s equation. The correct point is (1, 1).

   ---

   Concept & Intuition

   When a circle cuts the axes, the lengths of the intercepts are related to the centre’s coordinates. If the centre is $(h, k)$, the x-intercept length is $2\sqrt{r^2 - k^2}$ and the y-intercept length is $2\sqrt{r^2 - h^2}$.

   We also know the centre lies on the given diameter line $3x + 2y = 0$ and is in the 4th quadrant ($h > 0, k < 0$).

   We solve for $h, k, r$, then test each option.

   ---

   Step-by-step solution

   1. Set up intercept equations

      For a circle $(x - h)^2 + (y - k)^2 = r^2$:

      • x-intercept length = $2\sqrt{r^2 - k^2} = 2\sqrt{7}$ → $r^2 - k^2 = 7$
      • y-intercept length = $2\sqrt{r^2 - h^2} = 2\sqrt{12}$ → $r^2 - h^2 = 12$

   2. Use the diameter condition

      The centre lies on $3x + 2y = 0$ → $3h + 2k = 0$ → $k = -\frac{3}{2}h$.

      Since centre is in 4th quadrant, $h > 0$ and $k < 0$, consistent with $k = -\frac{3}{2}h$.

   3. Solve for $h$ and $r^2$

      From $r^2 - k^2 = 7$ and $k = -\frac{3}{2}h$:

   $r^2 - \frac{9}{4}h^2 = 7 \quad \text{(1)}$

   From $r^2 - h^2 = 12$:

   $r^2 - h^2 = 12 \quad \text{(2)}$

   Subtract (2) from (1):

   $\left(r^2 - \frac{9}{4}h^2\right) - (r^2 - h^2) = 7 - 12$

   $-\frac{9}{4}h^2 + h^2 = -5$

   $-\frac{5}{4}h^2 = -5 \quad\Rightarrow\quad h^2 = 4$

   So $h = 2$ (positive, since 4th quadrant). Then $k = -\frac{3}{2}(2) = -3$.

   4. Find $r^2$ From (2): $r^2 = 12 + h^2 = 12 + 4 = 16$. So the circle is $(x - 2)^2 + (y + 3)^2 = 16$. …

   🔒 Start your 14-day free trial to see the full solution →
7. TG EAPCET 2026Set eng-2026-05-10-AN1 markMCQ
   Q.If $(h, k)$ is the external centre of similitude for the circles $x^2 + y^2 - 6x - 10y + 9 = 0$ and $x^2 + y^2 + 6x + 6y + 2 = 0$, then $k - h =$ (A) $-\dfrac{7}{9}$ (B) $-8$ (C) $52$ (D) $\dfrac{1}{9}$

   ›Reveal solutionSolution

   The external centre of similitude divides the line joining the centres externally in the ratio of the radii. After finding centres and radii, the coordinates $(h,k)$ are computed, and $k-h$ simplifies to $-8$, so option (B) is correct.

   Concept & Intuition

   Two circles have two centres of similitude (internal and external). The external centre lies on the line joining the centres and divides it externally in the ratio of the radii. This is analogous to a “zoom” point from which the circles appear to have the same tangent lines. By finding the centres and radii, we can locate this point and then compute $k-h$.

   ---

   Step-by-step solution

   1. Rewrite each circle in standard form First circle:

   $x^2 + y^2 - 6x - 10y + 9 = 0$

   Complete squares:

   $(x^2 - 6x) + (y^2 - 10y) = -9$

   $(x-3)^2 - 9 + (y-5)^2 - 25 = -9$

   $(x-3)^2 + (y-5)^2 = 25$

   So centre $C_1 = (3,5)$ and radius $r_1 = 5$.

   Second circle:

   $x^2 + y^2 + 6x + 6y + 2 = 0$

   Complete squares:

   $(x^2 + 6x) + (y^2 + 6y) = -2$

   $(x+3)^2 - 9 + (y+3)^2 - 9 = -2$

   $(x+3)^2 + (y+3)^2 = 16$

   So centre $C_2 = (-3,-3)$ and radius $r_2 = 4$.

   2. External division formula For external centre of similitude $P$, the point divides $C_1C_2$ externally in the ratio $r_1 : r_2$. If $P$ divides $C_1$ and $C_2$ externally in ratio $m:n$, then

   $P = \left( \frac{m x_2 - n x_1}{m - n}, \frac{m y_2 - n y_1}{m - n} \right)$

   Here $m = r_1 = 5$, $n = r_2 = 4$. …

   🔒 Start your 14-day free trial to see the full solution →
8. TG EAPCET 2026Set eng-2026-05-10-AN1 markMCQ
   Q.A straight line $3x + y - 3 = 0$ cuts the parabola $x^2 - 2x + y + 1 = 0$ in two points. The area of the region enclosed by these two curves is (A) $\dfrac{9}{3}$ (B) $\dfrac{9}{2}$ (C) $\dfrac{25}{6}$ (D) $\dfrac{29}{6}$

   ›Reveal solutionSolution

   The area between a line and a parabola is found by integrating the difference of their $y$-expressions between their intersection points. After solving, the area is $\frac{9}{2}$, so the correct option is (B).

   ---

   Concept and Intuition

   We have two curves: a straight line and a parabola. The region enclosed by them is the area between the curves from one intersection point to the other. The standard method:

   1. Solve for intersection points (where their $y$-values are equal).
   2. Express both curves as $y = f(x)$ (or $x = g(y)$).
   3. Integrate the absolute difference of the functions over the $x$-interval (or $y$-interval) between intersections.

   Here, the line is already in a simple form, and the parabola is given in a form that can be rearranged to $y = -x^2 + 2x - 1$, which is a downward-opening parabola. The line lies above the parabola in the region between the two intersection points, so we integrate (line $-$ parabola) from left to right.

   ---

   Step-by-step solution

   1. Rewrite the parabola in standard form

   Given: $x^2 - 2x + y + 1 = 0$

   Solve for $y$:

   $y = -x^2 + 2x - 1$

   This is a downward parabola with vertex at $(1,0)$.

   2. Rewrite the line in slope-intercept form

   Given: $3x + y - 3 = 0$

   So:

   $y = -3x + 3$

   3. Find intersection points

   Set the $y$'s equal:

   $-3x + 3 = -x^2 + 2x - 1$

   Bring all terms to one side:

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

   $0 = -x^2 + 5x - 4$

   Multiply by $-1$:

   $x^2 - 5x + 4 = 0$

   Factor:

   $(x-1)(x-4) = 0$

   So $x = 1$ and $x = 4$.

   4. Determine which curve is on top

   Pick a test point between $x=1$ and $x=4$, say $x=2$:

   • Line: $y = -3(2)+3 = -3$
   • Parabola: $y = -(2)^2 + 2(2) - 1 = -4 + 4 - 1 = -1$

   At $x=2$, the parabola ($y=-1$) is above the line ($y=-3$). Wait — that means the parabola is above the line in this interval. Let’s check at $x=3$:

   • Line: $y = -9+3 = -6$
   • Parabola: $y = -9 + 6 - 1 = -4$

   Yes, parabola is above the line. So the region enclosed is between $x=1$ and $x=4$, with the parabola on top and the line on the bottom.

   5. Set up the area integral

   Area = $\int_{x=1}^{4} \big[ y_{\text{top}} - y_{\text{bottom}} \big] \, dx$

   $= \int_{1}^{4} \big[ (-x^2 + 2x - 1) - (-3x + 3) \big] \, dx$

   Simplify the integrand:

   $= \int_{1}^{4} \big( -x^2 + 2x - 1 + 3x - 3 \big) \, dx$

   $= \int_{1}^{4} \big( -x^2 + 5x - 4 \big) \, dx$

   6. Integrate

   $\int (-x^2 + 5x - 4) \, dx = -\frac{x^3}{3} + \frac{5x^2}{2} - 4x$

   Evaluate from $1$ to $4$:

   At $x=4$: …

   🔒 Start your 14-day free trial to see the full solution →
9. TG EAPCET 2026Set eng-2026-05-11-FN1 markMCQ
   Q.If the foci of the ellipse $\dfrac{x^2}{25} + \dfrac{y^2}{k^2} = 1$ $(k^2 < 25)$ and the hyperbola $\dfrac{x^2}{k} - \dfrac{y^2}{5} = 1$ are same then the product of the length of the latus rectum of the ellipse and that of hyperbola is (A) 25 (B) 50 (C) 16 (D) 32

   ›Reveal solutionSolution

   The condition that the foci coincide gives $k = 3$, from which the latus rectum lengths are computed and multiplied to get $32$.

   The problem gives an ellipse and a hyperbola whose foci are the same. That single condition will determine the unknown parameter $k$, and once we have $k$ we can compute the latus rectum of each conic and multiply them.

   The key idea: for both conics, the foci lie on the x‑axis (since the ellipse has its major axis along x and the hyperbola opens left‑right). Their foci are at $(\pm c, 0)$, where $c$ is the distance from the centre to a focus. For the ellipse, $c^2 = a^2 - b^2$; for the hyperbola, $c^2 = a^2 + b^2$. Setting these equal gives an equation in $k$.

   Let’s go step by step.

   1. Ellipse: $\frac{x^2}{25} + \frac{y^2}{k^2} = 1$, with $k^2 < 25$.

      Here $a^2 = 25$, $b^2 = k^2$. Since $a > b$, the major axis is along $x$.

      Focal distance: $c_e^2 = a^2 - b^2 = 25 - k^2$.

   2. Hyperbola: $\frac{x^2}{k} - \frac{y^2}{5} = 1$.

      This is of the form $\frac{x^2}{a_h^2} - \frac{y^2}{b_h^2} = 1$, so $a_h^2 = k$ and $b_h^2 = 5$.

      Focal distance: $c_h^2 = a_h^2 + b_h^2 = k + 5$.

   3. Foci are the same: $c_e^2 = c_h^2$.

   $25 - k^2 = k + 5$

   Rearranging: $k^2 + k - 20 = 0$.

   Factor: $(k + 5)(k - 4) = 0$, so $k = -5$ or $k = 4$.

   But $k^2 < 25$ from the ellipse condition, and $k$ appears as $a_h^2 = k$ in the hyperbola — it must be positive (since $a_h^2 > 0$). So $k = 4$ is the only valid choice.

   4. Latus rectum of the ellipse: For an ellipse $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ with $a > b$, the length is $\frac{2b^2}{a}$. …

   🔒 Start your 14-day free trial to see the full solution →
10. TG EAPCET 2026Set eng-2026-05-11-AN1 markMCQ
    Q.Equation of the image of the pair of lines $ax^2 + 2hxy + by^2 = 0$ with respect to $y = 0$ is (A) $ax^2 - 2hxy - by^2 = 0$ (B) $ax^2 - 2hxy + by^2 = 0$ (C) $ax^2 + 2hxy - by^2 = 0$ (D) $bx^2 - 2hxy + ay^2 = 0$

    ›Reveal solutionSolution

    Reflecting a pair of lines across the x‑axis (y = 0) changes the sign of any term containing an odd power of y. Since the given equation is homogeneous of degree 2, the image is obtained by replacing y with –y, which flips the sign of the xy‑term. The correct result is $ax^2 - 2hxy + by^2 = 0$, option (B).

    The key idea is that reflecting a curve across the line $y = 0$ (the x‑axis) means every point $(x, y)$ on the original curve is sent to $(x, -y)$ on the image. So to get the equation of the reflected curve, we simply replace $y$ by $-y$ in the original equation.

    The original equation is $ax^2 + 2hxy + by^2 = 0$. This is a homogeneous second‑degree equation representing a pair of straight lines through the origin. When we reflect these lines across the x‑axis, the resulting pair of lines is also through the origin, and its equation is obtained by the substitution $y \to -y$.

    Let’s do it step by step:

    1. Substitute $y \to -y$ Replace every $y$ in $ax^2 + 2hxy + by^2 = 0$ with $-y$:

    $a x^2 + 2h x (-y) + b (-y)^2 = 0.$

    2. Simplify each term

       • $a x^2$ is unchanged.
       • $2h x (-y) = -2hxy$.
       • $b (-y)^2 = b y^2$ because $(-y)^2 = y^2$.

    3. Write the resulting equation

    $a x^2 - 2hxy + b y^2 = 0.$

    That’s the equation of the reflected pair of lines. …

    🔒 Start your 14-day free trial to see the full solution →
11. TG EAPCET 2025Set eng-2025-05-02-AN1 markMCQ
    Q.O$(0,0)$, B$(-3,-1)$, C$(-1,-3)$ are vertices of a triangle OBC. D is a point on OC and E is a point on OB. If the equation of DE is $2x+2y+\sqrt{2}=0$, then the ratio in which the line DE divides the altitude of the triangle OBC is (A) $\sqrt{2}:4\sqrt{2}+2$ (B) $1:4\sqrt{2}+1$ (C) $\sqrt{2}:4\sqrt{2}-2$ (D) $1:4\sqrt{2}-1$

    ›Reveal solutionSolution

    The line DE divides the altitude from O to BC in the ratio $1 : 4\sqrt{2} - 1$, which corresponds to option (D). The key is to find the foot of the altitude, then the intersection of DE with that altitude, and compute the division ratio.

    ---

    Concept & Intuition

    We have triangle OBC with O at the origin. The altitude from O is the line through O perpendicular to BC. The line DE is given, and we want the ratio in which DE cuts that altitude. So we need:

    • Equation of BC.
    • Equation of altitude from O (call it $OH$, where $H$ is foot on BC).
    • Intersection point $P$ of DE and OH.
    • Ratio $OP : PH$ (or $OP : OH$ depending on interpretation — here it's the division of the altitude segment from O to BC).

    The altitude is a segment from O to BC; DE cuts it somewhere between O and H. We find distances along that line.

    ---

    Step-by-step

    1. Find equation of BC. $B = (-3, -1)$, $C = (-1, -3)$. Slope of BC:

    $m_{BC} = \frac{-3 - (-1)}{-1 - (-3)} = \frac{-2}{2} = -1.$

    Equation through B:

    $y + 1 = -1(x + 3) \quad\Rightarrow\quad y = -x - 4.$

    So BC: $x + y + 4 = 0$.

    2. Altitude from O to BC. Since BC has slope $-1$, a perpendicular line has slope $1$. Through O(0,0):

    $y = x.$

    This is the altitude line $OH$.

    3. Foot H of altitude on BC. Intersection of $y = x$ and $x + y + 4 = 0$:

    $x + x + 4 = 0 \quad\Rightarrow\quad 2x = -4 \quad\Rightarrow\quad x = -2,\; y = -2.$

    So $H = (-2, -2)$.

    Length $OH = \sqrt{(-2)^2 + (-2)^2} = \sqrt{8} = 2\sqrt{2}$.

    4. Intersection P of DE with altitude OH. DE: $2x + 2y + \sqrt{2} = 0$ simplifies to $x + y + \frac{\sqrt{2}}{2} = 0$. Altitude: $y = x$. Substitute:

    $x + x + \frac{\sqrt{2}}{2} = 0 \quad\Rightarrow\quad 2x = -\frac{\sqrt{2}}{2} \quad\Rightarrow\quad x = -\frac{\sqrt{2}}{4}.$

    So $P = \left(-\frac{\sqrt{2}}{4}, -\frac{\sqrt{2}}{4}\right)$.

    5. Distances along the altitude. …

    🔒 Start your 14-day free trial to see the full solution →
12. TG EAPCET 2025Set eng-2025-05-02-AN1 markMCQ
    Q.The locus of the centre of the circle touching the x-axis and passing through the point $(-1,1)$ is (A) a circle with centre at $\left(-1, \frac{1}{2}\right)$ (B) a pair of lines intersecting at $(-1,1)$ (C) a parabola with focus at $(-1,1)$ (D) a hyperbola with centre at $(-1,1)$

    ›Reveal solutionSolution

    The centre of a circle that touches the x‑axis and passes through a fixed point must satisfy that its distance to the fixed point equals its distance to the x‑axis (the y‑coordinate). This equality defines a parabola, and the given point becomes the focus. The correct option is (C).

    ---

    Concept & Intuition

    A circle that touches the x‑axis has its centre at some point $(h,k)$ with radius equal to $|k|$ (the perpendicular distance to the x‑axis). If it also passes through $(-1,1)$, then the distance from the centre to that point must equal the radius. Setting these distances equal gives a relation between $h$ and $k$ — and that relation is the equation of a parabola. The key geometric insight: the set of points equidistant from a fixed point (the given point) and a fixed line (the x‑axis) is a parabola, with the fixed point as focus and the fixed line as directrix.

    ---

    Step‑by‑step reasoning

    1. Set up variables Let the centre of the circle be $(h,k)$. Since the circle touches the x‑axis, the radius $r$ equals the vertical distance from the centre to the x‑axis:

    $r = |k|.$

    (We can assume $k \neq 0$; otherwise the circle would be degenerate.)

    2. Use the condition that the circle passes through $(-1,1)$ The distance from the centre to $(-1,1)$ must equal the radius:

    $\sqrt{(h + 1)^2 + (k - 1)^2} = |k|.$

    3. Square both sides (both sides are non‑negative)

    $(h + 1)^2 + (k - 1)^2 = k^2.$

    4. Simplify the equation Expand:

    $(h + 1)^2 + k^2 - 2k + 1 = k^2.$

    Cancel $k^2$ on both sides:

    $(h + 1)^2 - 2k + 1 = 0.$

    So:

    $(h + 1)^2 + 1 = 2k.$

    Hence:

    $k = \frac{(h + 1)^2 + 1}{2}.$

    5. Interpret the relation Replace $(h,k)$ by $(x,y)$ to get the locus:

    $y = \frac{(x + 1)^2 + 1}{2}.$

    This is a quadratic in $x$ — specifically a parabola opening upward.

    Rewrite in standard form:

    $y = \frac{1}{2}(x + 1)^2 + \frac{1}{2}.$

    Compare with $y = \frac{1}{4p}(x - x_0)^2 + y_0$: here $x_0 = -1$, $y_0 = \frac12$, and $\frac{1}{4p} = \frac12 \Rightarrow p = \frac12$. …

    🔒 Start your 14-day free trial to see the full solution →