Q.Find the centre and radius of the circle $(x + 5)^2 + (y - 3)^2 = 36$.

Showing the 12 most recent of 28 on this concept.

1. TG EAPCET 2026Set eng-2026-05-09-AN1 markMCQ
   Q.Let ABC be a triangle and $A = (-2, 3)$. If $7x - y + 2 = 0$ and $4x - 7y + 44 = 0$ are the equations of the medians drawn through the vertices B and C respectively, then AB = (A) $5\sqrt{2}$ (B) $3\sqrt{5}$ (C) $\sqrt{5}$ (D) $\dfrac{\sqrt{57}}{2}$

   ›Reveal solutionSolution

   To find the length of side AB, we first use the properties of medians to determine the coordinates of vertex B. A median connects a vertex to the midpoint of the opposite side, and the vertex itself lies on its respective median. By setting up and solving a system of equations based on these facts, we find the coordinates of B and then calculate the distance AB. The length of AB is $\boxed{3\sqrt{5}}$.

   Concept and Intuition

   In any triangle, a median is a line segment joining a vertex to the midpoint of the opposite side. The problem provides the coordinates of vertex A and the equations of two medians: one drawn from vertex B and another from vertex C. Our goal is to find the length of the side AB.

   To find the length of AB, we need the coordinates of both A and B. We are given $A = (-2, 3)$. Therefore, the main task is to find the coordinates of vertex B.

   Here's the intuition:

   1. Vertex on Median: Vertex B must lie on the median drawn from B. Similarly, vertex C must lie on the median drawn from C. This gives us two equations relating the coordinates of B and C.
   2. Midpoint on Median: The median from B connects B to the midpoint of AC. This midpoint must lie on the line representing the median from B. Similarly, the midpoint of AB must lie on the line representing the median from C. These two conditions will give us two more equations.
   3. System of Equations: We will have a system of four linear equations involving the coordinates of B and C. Solving this system will give us the coordinates of B (and C, though C is not strictly needed for the final answer).
   4. Distance Formula: Once we have the coordinates of A and B, we can use the distance formula to find the length of AB.

   Let $A = (x_A, y_A) = (-2, 3)$.

   Let $B = (x_B, y_B)$.

   Let $C = (x_C, y_C)$.

   The equation of the median from B is $L_B: 7x - y + 2 = 0$.

   The equation of the median from C is $L_C: 4x - 7y + 44 = 0$.

   Step-by-step Derivation

   1. Formulate equations based on vertices lying on their medians: Since vertex B lies on the median $L_B$, its coordinates $(x_B, y_B)$ must satisfy the equation of $L_B$:

   $7x_B - y_B + 2 = 0 \implies y_B = 7x_B + 2 \quad \text{(Equation 1)}$

   Similarly, since vertex C lies on the median $L_C$, its coordinates $(x_C, y_C)$ must satisfy the equation of $L_C$:
   

   $4x_C - 7y_C + 44 = 0 \implies 7y_C = 4x_C + 44 \implies y_C = \frac{4x_C + 44}{7} \quad \text{(Equation 2)}$

   2. Formulate equations based on midpoints lying on medians: The median from B connects B to the midpoint of AC. Let $M_{AC}$ be the midpoint of AC.

   $M_{AC} = \left(\frac{x_A + x_C}{2}, \frac{y_A + y_C}{2}\right) = \left(\frac{-2 + x_C}{2}, \frac{3 + y_C}{2}\right)$

   Since $M_{AC}$ lies on the median $L_B$, its coordinates must satisfy $7x - y + 2 = 0$:
   

   $7\left(\frac{-2 + x_C}{2}\right) - \left(\frac{3 + y_C}{2}\right) + 2 = 0$

   Multiply the entire equation by 2 to clear denominators:
   

   $7(-2 + x_C) - (3 + y_C) + 4 = 0$

   $-14 + 7x_C - 3 - y_C + 4 = 0$

   $7x_C - y_C - 13 = 0 \quad \text{(Equation 3)}$

   The median from C connects C to the midpoint of AB. Let $M_{AB}$ be the midpoint of AB.
   

   $M_{AB} = \left(\frac{x_A + x_B}{2}, \frac{y_A + y_B}{2}\right) = \left(\frac{-2 + x_B}{2}, \frac{3 + y_B}{2}\right)$

   Since $M_{AB}$ lies on the median $L_C$, its coordinates must satisfy $4x - 7y + 44 = 0$:
   

   $4\left(\frac{-2 + x_B}{2}\right) - 7\left(\frac{3 + y_B}{2}\right) + 44 = 0$

   Multiply the entire equation by 2:
   

   $4(-2 + x_B) - 7(3 + y_B) + 88 = 0$

   $-8 + 4x_B - 21 - 7y_B + 88 = 0$

   $4x_B - 7y_B + 59 = 0 \quad \text{(Equation 4)}$

   3. Solve the system of equations to find the coordinates of B: We now have a system of four equations:
      1. $y_B = 7x_B + 2$ …

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2. TG EAPCET 2026Set eng-2026-05-09-FN1 markMCQ
   Q.If two vertices of a quadrilateral are the centres of the circles $S \equiv x^2 + y^2 - 2x - 2y - 2 = 0$, $S' \equiv x^2 + y^2 - 6x - 6y + 14 = 0$ and the other two vertices of that quadrilateral are the points of intersection of these two circles $S = 0$ and $S' = 0$ then the area of the quadrilateral is (A) $4$ (B) $5\sqrt{2}$ (C) $7$ (D) $\dfrac{5}{\sqrt{2}}$

   ›Reveal solutionSolution

   Both circles have radius $2$; the line of centres ($2\sqrt2$) and the common chord ($2\sqrt2$) are perpendicular diagonals, so area $=\tfrac12(2\sqrt2)(2\sqrt2)=4$.

   Circles. $S:\;x^2+y^2-2x-2y-2=0$ has centre $C_1(1,1)$ and $r_1^2=1+1+2=4$, so $r_1=2$. $S':\;x^2+y^2-6x-6y+14=0$ has centre $C_2(3,3)$ and $r_2^2=9+9-14=4$, so $r_2=2$.

   Vertices. The quadrilateral is $C_1,\,A,\,C_2,\,B$, where $A,B$ are the two intersection points. Its diagonals are $C_1C_2$ (line of centres) and $AB$ (common chord); the line of centres is perpendicular to the common chord.

   Diagonal $C_1C_2$. $\;d=\sqrt{(3-1)^2+(3-1)^2}=\sqrt8=2\sqrt2.$ …

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3. TG EAPCET 2026Set eng-2026-05-10-FN1 markMCQ
   Q.If $x+y-2=0$ and $2x-y-1=0$ represent two adjacent sides of a parallelogram and $x+4y-14=0$ represents one of its diagonals then one of the vertices of the parallelogram is (A) $\left(0, \dfrac{7}{2}\right)$ (B) $(-2, 3)$ (C) $(-1, 6)$ (D) $(2, -4)$

   ›Reveal solutionSolution

   The intersection of the two given sides gives one vertex; the diagonal equation passes through the opposite pair of vertices, not this one. Solving the system of the diagonal with each side yields two more vertices, and the fourth vertex follows from the parallelogram's midpoint property. The correct vertex among the options is (C) $(-1, 6)$.

   The problem gives you two adjacent sides of a parallelogram and one of its diagonals. The key idea: in a parallelogram, the diagonals bisect each other. So if you find the vertex where the two given sides meet, that’s one corner. By checking whether the given diagonal passes through that corner, you can tell which diagonal it is; intersecting it with each side then gives the other two vertices, and the fourth vertex is determined by the midpoint condition.

   Let’s work it through.

   1. Find the vertex where the two adjacent sides meet.

      Solve the equations of the sides:

      $x + y - 2 = 0$ and $2x - y - 1 = 0$.

      Adding them: $(x + y - 2) + (2x - y - 1) = 0 \Rightarrow 3x - 3 = 0 \Rightarrow x = 1$.

      Substitute into $x + y - 2 = 0$: $1 + y - 2 = 0 \Rightarrow y = 1$.

      So one vertex is $A(1, 1)$.

   2. Determine which diagonal is given.

      In a parallelogram $ABCD$, the two diagonals are $AC$ and $BD$. Check whether the given diagonal passes through $A$: $1 + 4(1) - 14 = -9 \neq 0$, so $A$ does not lie on it. Since each diagonal connects a pair of opposite vertices, and $A$ is not on this line, the given diagonal $x + 4y - 14 = 0$ must be the diagonal $BD$ — connecting the two vertices adjacent to $A$, which we call $B$ and $D$.

   3. Find the intersection of the diagonal with each side.

      The diagonal meets side $AB$ (equation $x + y - 2 = 0$) at some point. Solve:

      $x + y - 2 = 0$ and $x + 4y - 14 = 0$.

      Subtract: $(x + 4y - 14) - (x + y - 2) = 0 \Rightarrow 3y - 12 = 0 \Rightarrow y = 4$.

      Then $x + 4 - 2 = 0 \Rightarrow x = -2$.

      So the diagonal meets side $AB$ at $(-2, 4)$. This point is $B$, so $B = (-2, 4)$.

      Next, the diagonal meets side $AD$ (equation $2x - y - 1 = 0$). Solve:

      $2x - y - 1 = 0$ and $x + 4y - 14 = 0$. …

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4. TG EAPCET 2026Set eng-2026-05-10-FN1 markMCQ
   Q.For the ellipse $\dfrac{x^2}{36} + \dfrac{y^2}{25} = 1$, the straight line $2x + y - 5 = 0$ is (A) a tangent (B) a normal (C) a focal chord (D) a chord not passing through its foci

   ›Reveal solutionSolution

   Substituting the line into the ellipse gives two distinct real intersection points, so it is a secant that does not pass through either focus.

   The ellipse is $\dfrac{x^2}{36}+\dfrac{y^2}{25}=1$, so $a^2=36,\ b^2=25$ and $c^2=a^2-b^2=11$, giving foci at $(\pm\sqrt{11},0)$.

   Put $y=5-2x$ from the line $2x+y-5=0$ into the ellipse:

   $\frac{x^2}{36}+\frac{(5-2x)^2}{25}=1.$

   Multiplying by $900$: $\;25x^2+36(5-2x)^2=900$, i.e. $169x^2-720x=0$, so

   $x=0 \quad\text{or}\quad x=\tfrac{720}{169}\approx 4.26.$ …

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5. TG EAPCET 2025Set eng-2025-05-02-FN1 markMCQ
   Q.If $\dfrac{2+3i}{i-2} - \dfrac{4i-3}{3+4i} = x+iy$, then $3x+y=$ (A) $4$ (B) $-4$ (C) $-2$ (D) $2$

   ›Reveal solutionSolution

   Simplify each fraction by multiplying by the conjugate of its denominator, combine the results, and extract the real and imaginary parts to find $x$ and $y$. The answer is $3x + y = \boxed{-2}$.

   When dividing complex numbers, the standard technique is to multiply numerator and denominator by the conjugate of the denominator. This converts the denominator into a real number (since $(a+bi)(a-bi) = a^2 + b^2$), making it straightforward to separate the real and imaginary parts.

   We need to simplify each fraction separately, then combine them.

   First fraction: $\dfrac{2+3i}{i-2}$

   1. Rewrite the denominator as $-2+i$ to see it more clearly. The conjugate of $-2+i$ is $-2-i$.

   2. Multiply numerator and denominator by $-2-i$:

   $\frac{2+3i}{-2+i} \cdot \frac{-2-i}{-2-i} = \frac{(2+3i)(-2-i)}{(-2+i)(-2-i)}$

   3. Expand the denominator:

   $(-2+i)(-2-i) = (-2)^2 - (i)^2 = 4 - (-1) = 5$

   4. Expand the numerator:

   $(2+3i)(-2-i) = 2(-2) + 2(-i) + 3i(-2) + 3i(-i)$

   $= -4 - 2i - 6i - 3i^2 = -4 - 8i + 3 = -1 - 8i$

   5. So the first fraction equals $\dfrac{-1-8i}{5} = -\dfrac{1}{5} - \dfrac{8}{5}i$.

   Second fraction: $\dfrac{4i-3}{3+4i}$

   6. The conjugate of $3+4i$ is $3-4i$.

   7. Multiply numerator and denominator by $3-4i$:

   $\frac{-3+4i}{3+4i} \cdot \frac{3-4i}{3-4i} = \frac{(-3+4i)(3-4i)}{(3+4i)(3-4i)}$

   8. Expand the denominator:

   $(3+4i)(3-4i) = 9 - (4i)^2 = 9 - 16(-1) = 9 + 16 = 25$

   9. Expand the numerator: $(-3+4i)(3-4i) = -3(3) + (-3)(-4i) + 4i(3) + 4i(-4i)$ …

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6. TG EAPCET 2025Set eng-2025-05-02-AN1 markMCQ
   Q.Suppose C1 and C2 are two circles having no common points, then (A) There will be 3 common tangents to C1 and C2 (B) There will be exactly two common tangents to C1 and C2 (C) There will be no common tangent or there will be exactly two common tangents to C1 and C2 (D) There will be no common tangents or there will be four common tangents to C1 and C2

   ›Reveal solutionSolution

   The number of common tangents between two circles depends entirely on their relative positions. For two circles with no common points, they are either completely separate (outside each other) or one lies entirely inside the other. In the first case there are 4 common tangents; in the second case there are 0. So the correct choice is (D).

   ---

   The Core Idea

   The number of common tangents to two circles is a geometric "fingerprint" of how the circles are placed relative to each other.

   • If circles intersect, they share internal and external tangents in specific counts.
   • If they have no common points, only two possibilities exist:
     1. One circle lies completely outside the other (they are separate).
     2. One circle lies completely inside the other (they are nested).

   These two cases give different numbers of common tangents, so the answer cannot be a single fixed number.

   ---

   Step-by-Step Reasoning

   1. Recall the classification of relative positions of two circles

      Let the radii be $r_1$ and $r_2$ (with $r_1 \ge r_2$) and let $d$ be the distance between their centers.

      • If $d > r_1 + r_2$: circles are separate (no intersection).
      • If $d = r_1 + r_2$: circles touch externally (one common point).
      • If $|r_1 - r_2| < d < r_1 + r_2$: circles intersect at two points.
      • If $d = |r_1 - r_2|$: circles touch internally (one common point).
      • If $d < |r_1 - r_2|$: one circle lies completely inside the other (no common points).

   2. Focus on the "no common points" condition

      The problem states that C1 and C2 have no common points. This eliminates the cases where they touch or intersect. So only two subcases remain:

      • Case A: $d > r_1 + r_2$ (circles are separate).
      • Case B: $d < |r_1 - r_2|$ (one circle is inside the other).

   3. Count common tangents for Case A (separate circles)

      When two circles lie completely outside each other, you can draw:

      • Two direct (external) common tangents — these do not cross the line segment joining the centers.
      • Two transverse (internal) common tangents — these cross between the circles. That gives a total of 4 common tangents.

   4. Count common tangents for Case B (one circle inside the other) …

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7. TG EAPCET 2025Set eng-2025-05-03-AN1 markMCQ
   Q.A straight line passing through a point (3,2) cuts X and Y-axes at the points A and B respectively. If a point P divides AB in the ratio 2:3, then the equation of the locus of point P is (A) $\frac{9}{x} + \frac{4}{y} = 1$ (B) $9x + 4y = 5xy$ (C) $4x + 9y = 5xy$ (D) $\frac{4}{x} + \frac{9}{y} = 1$

   ›Reveal solutionSolution

   The key idea is to express the intercepts of the variable line in terms of the coordinates of the dividing point P, then eliminate the parameters using the condition that the line passes through (3,2). The locus is $4x + 9y = 5xy$, which corresponds to option (C).

   Concept & Intuition

   We have a line that always goes through the fixed point (3,2). It cuts the x-axis at A and the y-axis at B. A point P divides AB in the ratio 2:3 (from A to B). As the line rotates about (3,2), P traces a curve — its locus.

   The natural approach: let the intercepts be $A(a,0)$ and $B(0,b)$. Then find P in terms of $a$ and $b$ using the section formula. Then use the fact that (3,2) lies on the line $\frac{x}{a} + \frac{y}{b} = 1$ to relate $a$ and $b$. Finally eliminate $a$ and $b$ to get a relation between the coordinates of P.

   Step-by-step solution

   1. Set up intercepts and the line equation Let the line meet the x-axis at $A(a,0)$ and the y-axis at $B(0,b)$, with $a \neq 0$, $b \neq 0$. The equation of this line in intercept form is

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

   2. Use the given fixed point The line passes through $(3,2)$, so substitute:

   $\frac{3}{a} + \frac{2}{b} = 1. \qquad(1)$

   3. Find coordinates of point P dividing AB in ratio 2:3 The point P divides segment AB from A to B in the ratio $2:3$. Using the section formula (internal division):

   $P = \left( \frac{2 \cdot 0 + 3 \cdot a}{2+3}, \frac{2 \cdot b + 3 \cdot 0}{2+3} \right) = \left( \frac{3a}{5}, \frac{2b}{5} \right).$

   Let the coordinates of P be $(x,y)$. Then

   $x = \frac{3a}{5}, \quad y = \frac{2b}{5}.$

   4. Express $a$ and $b$ in terms of $x$ and $y$ From the above:

   $a = \frac{5x}{3}, \quad b = \frac{5y}{2}.$

   5. Substitute into equation (1)

   $\frac{3}{\frac{5x}{3}} + \frac{2}{\frac{5y}{2}} = 1.$

   Simplify each term:

   $$$$

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8. TG EAPCET 2025Set eng-2025-05-03-FN1 markMCQ
   Q.If a complex number $z = x + iy$ represents a point P$(x, y)$ in the Argand plane and $z$ satisfies the condition that the imaginary part of $\dfrac{z-3}{z+3i}$ is zero, then the locus of the point P is (A) $x + y + 3 = 0,\ (x, y) \neq (0, -3)$ (B) $x - y - 3 = 0,\ (x, y) \neq (0, -3)$ (C) $2xy - 3x + 3y + 9 = 0,\ (x, y) \neq (0, -3)$ (D) $x^2 + y^2 - 3x + 3y = 0,\ (x, y) \neq (0, -3)$

   ›Reveal solutionSolution

   $\dfrac{z-3}{z+3i}$ is real when its imaginary part vanishes, giving the line $x-y-3=0$, with $(0,-3)$ excluded. Option (B).

   Solution

   Let $z=x+iy$, so

   $\frac{z-3}{z+3i}=\frac{(x-3)+iy}{x+i(y+3)}.$

   For a quotient $\dfrac{N}{D}$, $\ \operatorname{Im}\!\left(\dfrac{N}{D}\right)=\dfrac{\operatorname{Im}(N)\operatorname{Re}(D)-\operatorname{Re}(N)\operatorname{Im}(D)}{|D|^{2}}$.

   Here $\operatorname{Re}(N)=x-3,\ \operatorname{Im}(N)=y,\ \operatorname{Re}(D)=x,\ \operatorname{Im}(D)=y+3$, so the imaginary part of the numerator is

   $y\cdot x-(x-3)(y+3)=xy-\bigl(xy+3x-3y-9\bigr)=-3x+3y+9.$

   Setting this to zero:

   $-3x+3y+9=0\ \Longrightarrow\ x-y-3=0.$ …

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9. TG EAPCET 2025Set eng-2025-05-03-FN1 markMCQ
   Q.A straight line passing through a point (3, 2) cuts X and Y-axes at the points A and B respectively. If a point P divides AB in the ratio 2 : 3, then the equation of the locus of point P is (A) $9x + 4y = 5xy$ (B) $\dfrac{4}{x} + \dfrac{9}{y} = 1$ (C) $4x + 9y = 5xy$ (D) $\dfrac{9}{x} + \dfrac{4}{y} = 1$

   ›Reveal solutionSolution

   Write the line in intercept form, express P by the section formula, and eliminate the intercepts. The locus is $4x + 9y = 5xy$, option (C).

   Let the line meet the X-axis at $A=(a,0)$ and the Y-axis at $B=(0,b)$. In intercept form it is

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

   Since it passes through $(3,2)$,

   $\frac{3}{a} + \frac{2}{b} = 1. \quad (\star)$

   P divides $AB$ internally in the ratio $2:3$ (so $AP:PB = 2:3$). By the section formula, with $A=(a,0)$ and $B=(0,b)$,

   $x = \frac{3a + 2\cdot 0}{5} = \frac{3a}{5}, \qquad y = \frac{3\cdot 0 + 2b}{5} = \frac{2b}{5}.$

   Hence …

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10. TG EAPCET 2025Set eng-2025-05-04-AN1 markMCQ
    Q.Two families of lines are given by $ax + by + c = 0$ and $4a^2 + 9b^2 - c^2 - 12ab = 0$. Then the line common to both the families is (A) a line passing through $(-1,2)$ and $(2,3)$ (B) a line passing through $(3,2)$ and $(2,3)$ (C) a line passing through $(-3,-2)$ and $(-2,-3)$ (D) a line passing through $(2,-3)$ and $(-2,3)$

    ›Reveal solutionSolution

    Factoring the coefficient condition splits the lines into two pencils through $(-2,3)$ and $(2,-3)$; the line common to both is the one joining these points, i.e. option (D).

    Factor the condition.

    $4a^2 - 12ab + 9b^2 - c^2 = (2a-3b)^2 - c^2 = (2a-3b-c)(2a-3b+c)=0,$

    so either $c = 2a-3b$ or $c = -2a+3b$.

    Interpret each case in $ax+by+c=0$.

    • $c=2a-3b:\quad a x + b y + 2a - 3b = 0 \;\Rightarrow\; a(x+2) + b(y-3)=0$ — a pencil of lines through $(-2,3)$.
    • $c=-2a+3b:\quad a x + b y - 2a + 3b = 0 \;\Rightarrow\; a(x-2) + b(y+3)=0$ — a pencil of lines through $(2,-3)$. …

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11. TG EAPCET 2025Set eng-2025-05-04-AN1 markMCQ
    Q.The normal at a point on the parabola $y^2 = 4x$ passes through a point P. Two more normals to this parabola also pass through P. If the centroid of the triangle formed by the feet of these three normals is $G(2,0)$, then the abscissa of P is (A) $4$ (B) $-4$ (C) $5$ (D) $-5$

    ›Reveal solutionSolution

    The key idea is that for a cubic in the slope parameter $m$ of normals to $y^2=4x$, the centroid of the feet of three concurrent normals is fixed at $(2,0)$; this forces the sum of the slopes to be zero, which then determines the abscissa of P as $5$.

    ---

    We are given the parabola $y^2 = 4x$. Its standard form is $y^2 = 4ax$ with $a = 1$.

    A normal at a point $(t^2, 2t)$ (parameter $t$) has slope $-t$ and equation:

    $y = -t x + 2t + t^3.$

    If three distinct normals pass through a point $P(h, k)$, then the slopes $-t_1, -t_2, -t_3$ (or equivalently the parameters $t_1, t_2, t_3$) satisfy the same equation when we substitute $(h, k)$:

    $k = -t h + 2t + t^3 \quad\Rightarrow\quad t^3 + (2 - h)t - k = 0.$

    Thus $t_1, t_2, t_3$ are the three roots of the cubic:

    $t^3 + (2 - h)t - k = 0.$

    ---

    1. Relate the centroid of the feet to the roots. The feet of the normals are the points $(t_i^2, 2t_i)$. Their centroid $G$ is:

    $G = \left( \frac{t_1^2 + t_2^2 + t_3^2}{3},\; \frac{2(t_1 + t_2 + t_3)}{3} \right).$

    We are told $G = (2, 0)$. Hence:

    $\frac{2(t_1 + t_2 + t_3)}{3} = 0 \quad\Rightarrow\quad t_1 + t_2 + t_3 = 0.$

    And:

    $\frac{t_1^2 + t_2^2 + t_3^2}{3} = 2 \quad\Rightarrow\quad t_1^2 + t_2^2 + t_3^2 = 6.$

    2. Use the cubic’s coefficients. For the cubic $t^3 + 0\cdot t^2 + (2 - h)t - k = 0$, Vieta’s formulas give:

    $t_1 + t_2 + t_3 = 0,$

    $t_1 t_2 + t_2 t_3 + t_3 t_1 = 2 - h,$

    $t_1 t_2 t_3 = k.$

    The sum condition is already satisfied — consistent.

    3. Find $h$ from the sum of squares. We know:

    $t_1^2 + t_2^2 + t_3^2 = (t_1 + t_2 + t_3)^2 - 2(t_1 t_2 + t_2 t_3 + t_3 t_1).$

    Substituting:

    $$$$

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12. TG EAPCET 2025Set eng-2025-05-04-FN1 markMCQ
    Q.Two families of lines are given by $ax + by + c = 0$ and $4a^2 + 9b^2 - c^2 - 12ab = 0$. Then the line common to both the families is (A) a line passing through $(-3, -2)$ and $(-2, -3)$ (B) a line passing through $(-1, 2)$ and $(2, 3)$ (C) a line passing through $(2, -3)$ and $(-2, 3)$ (D) a line passing through $(3, 2)$ and $(2, 3)$

    ›Reveal solutionSolution

    The constraint factors as $(2a-3b)^{2}=c^{2}$, splitting the lines into two pencils through $(-2,3)$ and $(2,-3)$; the line common to both passes through those points — option (C).

    The coefficients satisfy $4a^{2}+9b^{2}-c^{2}-12ab=0$. Since $4a^{2}-12ab+9b^{2}=(2a-3b)^{2}$,

    $(2a-3b)^{2}-c^{2}=0\ \Rightarrow\ (2a-3b-c)(2a-3b+c)=0,$

    so $c=2a-3b$ or $c=3b-2a$. Each factor gives a family (pencil) of lines through a fixed point:

    • $c=2a-3b$: $a(x+2)+b(y-3)=0$, which passes through $(-2,3)$ for all $a,b$;
    • $c=3b-2a$: $a(x-2)+b(y+3)=0$, which passes through $(2,-3)$ for all $a,b$. …

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