Q.The value of $\lambda$ for which the vectors $3\hat{i}-6\hat{j}+\hat{k}$ and $2\hat{i}-4\hat{j}+\lambda\hat{k}$ are parallel is
(A) $\dfrac{2}{3}$
(B) $\dfrac{3}{2}$
(C) $\dfrac{5}{2}$
(D) $\dfrac{2}{5}$

Showing the 12 most recent of 69 on this concept.

1. TG EAPCET 2026Set eng-2026-05-09-AN1 markMCQ
   Q.Let $L_1 \equiv 3x + 4y - 1 = 0$, $L_2 \equiv 8x - 6y + 1 = 0$, $L_3 \equiv 12x + 9y - 1 = 0$ are three tangents drawn to the circle $x^2 + y^2 + 2gx + 2fy + c = 0$ and let $L_1 > 0$, $L_2 > 0$, $L_3 > 0$ for $x = -g$ and $y = -f$. Then $g + 2f =$ (A) $0$ (B) $\dfrac{1}{4}$ (C) $1$ (D) $\dfrac{1}{2}$

   ›Reveal solutionSolution

   Because each line is tangent to the circle, the centre $(-g,-f)$ is equidistant from all three lines, and the conditions $L_1,L_2,L_3>0$ at the centre fix the common (positive) signed distance. Equating the perpendicular distances and solving gives $g=\frac{19}{36}$, $f=-\frac{5}{36}$, so $g+2f=\frac14$. The correct option is (B).

   Concept & Intuition

   Three lines tangent to a circle form a triangle, and the circle's centre is the incentre of that triangle — the unique interior point equidistant from all three sides. The centre of the given circle is $(-g,-f)$. The conditions $L_1>0,\,L_2>0,\,L_3>0$ at $(-g,-f)$ say the centre lies on the positive side of each line, so its (signed) perpendicular distance to every line is the same positive value $r$ (the radius).

   Step-by-step

   1. Perpendicular distance to each line. For a line $ax+by+c=0$, the signed distance from a point is $\dfrac{ax+by+c}{\sqrt{a^2+b^2}}$. With the centre written as $(-g,-f)$:

   $\frac{-3g-4f-1}{5}=\frac{-8g+6f+1}{10}=\frac{-12g-9f-1}{15}=r>0,$

   using $\sqrt{3^2+4^2}=5$, $\sqrt{8^2+6^2}=10$, $\sqrt{12^2+9^2}=15$.

   2. Equate the first two distances.

   $\frac{-3g-4f-1}{5}=\frac{-8g+6f+1}{10}\ \Rightarrow\ 2(-3g-4f-1)=-8g+6f+1,$

   $-6g-8f-2=-8g+6f+1\ \Rightarrow\ 2g-14f=3\ \Rightarrow\ g-7f=\tfrac32.\qquad(1)$

   3. Equate the first and third distances.

   $\frac{-3g-4f-1}{5}=\frac{-12g-9f-1}{15}\ \Rightarrow\ 3(-3g-4f-1)=-12g-9f-1,$

   $$$$

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2. TG EAPCET 2026Set eng-2026-05-09-AN1 markMCQ
   Q.$T_1, T_2$ are the points of contact of a transverse common tangent drawn to the circles $x^2 + y^2 + 4x - 10y + 4 = 0$ and $x^2 + y^2 - 6x + 8y + 9 = 0$. If $T_1T_2$ is a horizontal line, then the midpoint of the line segment $T_1T_2$ is (A) $\begin{pmatrix} \dfrac{23}{10} \\ 0 \end{pmatrix}$ (B) $\begin{pmatrix} \dfrac{13}{10} \\ 0 \end{pmatrix}$ (C) $\begin{pmatrix} \dfrac{1}{2} \\ 0 \end{pmatrix}$ (D) $\begin{pmatrix} \dfrac{2}{5} \\ 0 \end{pmatrix}$

   ›Reveal solutionSolution

   The horizontal transverse tangent is the $x$-axis $y = 0$; the contact points are $(-2,0)$ and $(3,0)$, so the midpoint is $\left(\tfrac{1}{2},\,0\right)$.

   The two circles.

   • $C_1:\ x^2 + y^2 + 4x - 10y + 4 = 0$ — centre $(-2,5)$, radius $\sqrt{4 + 25 - 4} = 5$.
   • $C_2:\ x^2 + y^2 - 6x + 8y + 9 = 0$ — centre $(3,-4)$, radius $\sqrt{9 + 16 - 9} = 4$.

   A transverse (internal) common tangent has the two centres on opposite sides of it. Since the tangent is horizontal, write it as $y = c$.

   Distance conditions.

   • From $C_1$: $|5 - c| = 5 \Rightarrow c = 0$ or $c = 10$.
   • From $C_2$: $|-4 - c| = 4 \Rightarrow c = 0$ or $c = -8$. …

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3. TG EAPCET 2026Set eng-2026-05-10-AN1 markMCQ
   Q.For the circle $x^2 + y^2 + 12x - 4y - 9 = 0$, the straight line given by the equation $x + 2y - 3 = 0$ represents (A) a tangent (B) a chord but not its diameter (C) a diameter (D) a polar but not a tangent or a chord

   ›Reveal solutionSolution

   The perpendicular distance from the centre ($\sqrt5$) is less than the radius ($7$) and the centre is not on the line, so the line is a chord that is not a diameter. Correct option: (B).

   Centre and radius of the circle.

   For $x^2+y^2+12x-4y-9=0$: $g=6,\,f=-2,\,c=-9$.

   $\text{Centre }(-6,\,2),\qquad r=\sqrt{g^2+f^2-c}=\sqrt{36+4+9}=\sqrt{49}=7.$

   Distance from the centre to the line $x+2y-3=0$:

   $d=\frac{|(-6)+2(2)-3|}{\sqrt{1^2+2^2}}=\frac{|-5|}{\sqrt5}=\frac{5}{\sqrt5}=\sqrt5\approx2.24.$ …

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4. TG EAPCET 2026Set eng-2026-05-10-AN1 markMCQ
   Q.Let $L_1$ be a focal chord of the ellipse $9x^2 + 25y^2 = 225$ which makes equal intercepts on the coordinate axes. If $L_2$ is the tangent drawn at one end of a latus rectum that lies in the first quadrant, then the point of intersection of $L_1$ and $L_2$ lies (A) in the 1st quadrant (B) in the 2nd quadrant (C) on X-axis (D) on Y-axis

   ›Reveal solutionSolution

   The key idea is to find the equations of the two lines — one being a focal chord with equal intercepts (slope -1), the other being the tangent at the first-quadrant latus rectum endpoint — then solve for their intersection and determine its quadrant. The intersection lies on the Y-axis.

   ---

   Concept & Intuition

   We have an ellipse and two special lines.

   • L₁ is a focal chord (passes through a focus) and makes equal intercepts on the axes — that means its equation is of the form $x + y = c$ (since equal intercepts imply slope $-1$).
   • L₂ is the tangent at one end of a latus rectum in the first quadrant. The latus rectum is the chord through a focus perpendicular to the major axis. Its endpoints have known coordinates.
   • The intersection of these two lines will be a point; we just need to see which quadrant (or axis) it falls on.

   ---

   Step-by-step solution

   1. Rewrite the ellipse in standard form

   $9x^2 + 25y^2 = 225 \quad\Rightarrow\quad \frac{x^2}{25} + \frac{y^2}{9} = 1$

   So $a^2 = 25$, $b^2 = 9$. Hence $a = 5$, $b = 3$.

   2. Find the foci

      For an ellipse, $c^2 = a^2 - b^2 = 25 - 9 = 16$, so $c = 4$.

      The foci are at $(\pm 4, 0)$.

   3. Equation of L₁

      L₁ is a focal chord (passes through a focus) and makes equal intercepts on axes.

      A line with equal intercepts has equation $\frac{x}{k} + \frac{y}{k} = 1$ or $x + y = k$.

      Since it passes through a focus, say $(4,0)$, we get $4 + 0 = k \Rightarrow k = 4$.

      So L₁: $x + y = 4$.

      Tip

      “Equal intercepts” always means slope $-1$ (unless the line passes through the origin, which it doesn’t here). So the form is $x + y = \text{constant}$.

   4. Find the latus rectum endpoint in the first quadrant

      The latus rectum through focus $(4,0)$ is vertical (since major axis is horizontal). Its endpoints have $x = 4$ and $y$ satisfying the ellipse:

   $\frac{16}{25} + \frac{y^2}{9} = 1 \quad\Rightarrow\quad \frac{y^2}{9} = \frac{9}{25} \quad\Rightarrow\quad y = \pm \frac{9}{5}$

   In the first quadrant, $y > 0$, so the point is $P\left(4, \frac{9}{5}\right)$.

   5. Equation of L₂ (tangent at P) The tangent to the ellipse $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ at $(x_1, y_1)$ is:

   $\frac{x x_1}{a^2} + \frac{y y_1}{b^2} = 1$

   Substituting $x_1 = 4$, $y_1 = \frac{9}{5}$, $a^2 = 25$, $b^2 = 9$:

   $$$$

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5. TG EAPCET 2026Set eng-2026-05-10-FN1 markMCQ
   Q.The straight line given by the equation $\vec{r} = (4\vec{i} + 5\vec{j} + \vec{k}) + s(4\vec{i} + 6\vec{j} + 2\vec{k})$ is coplanar with a straight line given below. Choose the correct option (A) $\vec{r} = (\vec{i} - 2\vec{j} + 3\vec{k}) + p(2\vec{i} + 3\vec{j} - 4\vec{k})$ (B) $\vec{r} = (3\vec{i} - 4\vec{j} + 3\vec{k}) + q(-4\vec{i} + 5\vec{j} - 6\vec{k})$ (C) $\vec{r} = (2\vec{i} + 5\vec{j} - 4\vec{k}) + r(\vec{i} + 4\vec{j} - 3\vec{k})$ (D) $\vec{r} = (-4\vec{i} + 4\vec{j} + 4\vec{k}) + t(7\vec{i} + 5\vec{j})$

   ›Reveal solutionSolution

   Two lines are coplanar if the scalar triple product of their direction vectors and the vector joining a point on each line is zero. For the given line, only option (D) satisfies this condition.

   Concept & Intuition

   Two lines in 3D are coplanar if they lie in the same plane. This happens either when they are parallel (direction vectors are scalar multiples) or when they intersect. But there’s a third, more general case: they can be skew (not coplanar) or coplanar without intersecting (parallel but distinct). The universal test: take a point on each line, form the vector between them, and check if the three vectors (the two direction vectors and the connecting vector) are linearly dependent — i.e., their scalar triple product is zero. Geometrically, this means the volume of the parallelepiped they span is zero, so they all lie in a plane.

   Step-by-step solution

   1. Identify the given line

      Line $L_0$: $\vec{r} = (4\vec{i} + 5\vec{j} + \vec{k}) + s(4\vec{i} + 6\vec{j} + 2\vec{k})$

      Point $P_0 = (4,5,1)$, direction $\vec{d}_0 = (4,6,2)$.

      Notice $\vec{d}_0 = 2(2,3,1)$, so its direction is essentially $(2,3,1)$.

   2. Coplanarity condition

      For another line $L$: $\vec{r} = \vec{a} + t\vec{d}$, with point $P$ and direction $\vec{d}$, the lines are coplanar iff

   $(\vec{d}_0 \times \vec{d}) \cdot (\vec{a} - \vec{P}_0) = 0$

   That is, the scalar triple product $[\vec{d}_0, \vec{d}, \vec{P}_0\vec{P}] = 0$.

   3. Test each option

      Option (A):

      $P = (1,-2,3)$, $\vec{d} = (2,3,-4)$

      $\vec{d}_0 \times \vec{d} = \begin{vmatrix} \vec{i} & \vec{j} & \vec{k} \\ 4 & 6 & 2 \\ 2 & 3 & -4 \end{vmatrix} = \vec{i}(6\cdot(-4)-2\cdot3) - \vec{j}(4\cdot(-4)-2\cdot2) + \vec{k}(4\cdot3-6\cdot2)$

      $= \vec{i}(-24-6) - \vec{j}(-16-4) + \vec{k}(12-12) = (-30, 20, 0)$

      $\vec{a} - \vec{P}_0 = (1-4, -2-5, 3-1) = (-3, -7, 2)$

      Dot product: $(-30)(-3) + (20)(-7) + 0\cdot2 = 90 - 140 = -50 \neq 0$ → Not coplanar.

      Option (B):

      $P = (3,-4,3)$, $\vec{d} = (-4,5,-6)$

      $\vec{d}_0 \times \vec{d} = \begin{vmatrix} \vec{i} & \vec{j} & \vec{k} \\ 4 & 6 & 2 \\ -4 & 5 & -6 \end{vmatrix} = \vec{i}(6\cdot(-6)-2\cdot5) - \vec{j}(4\cdot(-6)-2\cdot(-4)) + \vec{k}(4\cdot5-6\cdot(-4))$

      $= \vec{i}(-36-10) - \vec{j}(-24+8) + \vec{k}(20+24) = (-46, 16, 44)$

      $\vec{a} - \vec{P}_0 = (3-4, -4-5, 3-1) = (-1, -9, 2)$

      Dot: $(-46)(-1) + (16)(-9) + (44)(2) = 46 - 144 + 88 = -10 \neq 0$ → Not coplanar.

      Option (C):

      $P = (2,5,-4)$, $\vec{d} = (1,4,-3)$

      Cross: $(4,6,2) \times (1,4,-3)$ …

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6. TG EAPCET 2026Set eng-2026-05-10-FN1 markMCQ
   Q.If the straight line $x + by + 1 = 0$ is a polar with respect to the circle $x^2 + y^2 - 8x + 10y - 8 = 0$, but not a tangent and not a chord then all the values of $b$ lie in the interval (A) $(-\infty, -1)$ (B) $\left(-\infty, -\frac{4}{3}\right) \cup \left(-\frac{3}{4}, \infty\right)$ (C) $(-1, 7)$ (D) $\left(-\frac{4}{3}, -\frac{3}{4}\right)$

   ›Reveal solutionSolution

   For the polar to miss the circle (no tangent, no chord) the pole must be interior, giving $12b^2+25b+12<0$, i.e. $b\in\left(-\tfrac43,-\tfrac34\right)$.

   The circle $x^2+y^2-8x+10y-8=0$ has centre $(4,-5)$ and radius $\sqrt{16+25+8}=7$.

   Any line is the polar of its pole. It is a tangent when the pole lies on the circle, and a chord (chord of contact) when the pole lies outside. It is neither when the pole is inside the circle — equivalently, the polar line does not meet the circle, so its distance from the centre exceeds the radius:

   $\frac{|4+b(-5)+1|}{\sqrt{1+b^2}}>7\ \Rightarrow\ |5-5b|>7\sqrt{1+b^2}.$

   Squaring: …

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7. TG EAPCET 2026Set eng-2026-05-11-FN1 markMCQ
   Q.If the eccentricity of an ellipse $\dfrac{x^2}{a^2} + \dfrac{y^2}{b^2} = 1$ $(a > b)$ is $e = \dfrac{\sqrt{3}}{2}$ and the equation of one of its directrices is $\sqrt{3}x - 4 = 0$, then $ab =$ (A) $a$ (B) $b$ (C) $a^2 - b^2$ (D) $\dfrac{e}{a}$

   ›Reveal solutionSolution

   By using the given eccentricity and directrix equation, we find the semi-major axis $a=2$ and semi-minor axis $b=1$. The product $ab$ is then $2$, which matches option (A).

   The problem asks us to find the product $ab$ for an ellipse, given its eccentricity and the equation of one of its directrices. The key idea is to use the standard relationships between the ellipse's parameters ($a$, $b$, $e$) and its directrices.

   For an ellipse given by $\dfrac{x^2}{a^2} + \dfrac{y^2}{b^2} = 1$ where $a > b$, the major axis lies along the x-axis.

   • The eccentricity $e$ is defined by the relation $b^2 = a^2(1-e^2)$.
   • The equations of the directrices are $x = \pm \dfrac{a}{e}$.

   We will use the given eccentricity and the directrix equation to find the value of $a$. Once $a$ is known, we can use the eccentricity relation to find $b$. Finally, we will calculate the product $ab$.

   1. Identify the given information and standard forms:

      The ellipse is given by $\dfrac{x^2}{a^2} + \dfrac{y^2}{b^2} = 1$ with $a > b$. This means the major axis is along the x-axis.

      The eccentricity is given as $e = \dfrac{\sqrt{3}}{2}$.

      The equation of one directrix is $\sqrt{3}x - 4 = 0$.

   2. Determine the standard form of the directrix equation:

      Rearrange the given directrix equation to match the standard form $x = \text{constant}$:

      $\sqrt{3}x - 4 = 0$

      $\sqrt{3}x = 4$

      $x = \dfrac{4}{\sqrt{3}}$

      For an ellipse with its major axis along the x-axis, the equations of the directrices are $x = \pm \dfrac{a}{e}$. We can equate the given directrix with one of these standard forms. Let's use $x = \dfrac{a}{e}$.

      $\dfrac{a}{e} = \dfrac{4}{\sqrt{3}}$

   3. Calculate the semi-major axis 'a':

      Substitute the given value of eccentricity $e = \dfrac{\sqrt{3}}{2}$ into the equation from the previous step:

      $\dfrac{a}{\left(\dfrac{\sqrt{3}}{2}\right)} = \dfrac{4}{\sqrt{3}}$

      $\dfrac{2a}{\sqrt{3}} = \dfrac{4}{\sqrt{3}}$

      Multiply both sides by $\sqrt{3}$:

      $2a = 4$

      $a = 2$

   4. Calculate the semi-minor axis 'b': …

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8. TG EAPCET 2025Set eng-2025-05-02-FN1 markMCQ
   Q.If the points $A(2,3),\ B(3,2)$ form a triangle with a variable point $p(t, t^2)$, where $t$ is a parameter, then the equation of the locus of the centroid of triangle $ABC$ is (A) $9x^2 - 30x - 3y + 20 = 0$ (B) $3x^2 - 10x - y + 10 = 0$ (C) $9y^2 - 30y - 3x + 20 = 0$ (D) $3y^2 - 10y - x + 10 = 0$

   ›Reveal solutionSolution

   The centroid of triangle ABC is the average of the three vertices. Substituting the coordinates of A, B, and the variable point P(t, t²) gives the centroid in terms of t. Eliminating t yields a quadratic relation between x and y, which matches option (B).

   Concept & Intuition

   The centroid of a triangle is simply the arithmetic mean of its vertices’ coordinates. When one vertex moves along a known curve (here the parabola $y = x^2$), the centroid traces its own path. The trick is to express the centroid’s coordinates in terms of the parameter $t$, then eliminate $t$ to find the Cartesian equation of that path.

   ---

   1. Write the centroid coordinates For points $A(2,3)$, $B(3,2)$, and $P(t, t^2)$, the centroid $G(x,y)$ is

   $x = \frac{2 + 3 + t}{3} = \frac{5 + t}{3}, \qquad y = \frac{3 + 2 + t^2}{3} = \frac{5 + t^2}{3}.$

   2. Eliminate the parameter $t$ From $x = \frac{5 + t}{3}$, solve for $t$:

   $t = 3x - 5.$

   Substitute into $y = \frac{5 + t^2}{3}$:

   $y = \frac{5 + (3x - 5)^2}{3}.$

   3. Simplify to a Cartesian equation Expand:

   $y = \frac{5 + (9x^2 - 30x + 25)}{3} = \frac{9x^2 - 30x + 30}{3}.$

   Divide through:

   $y = 3x^2 - 10x + 10.$

   Rearranging:

   $$3x^2 - 10x - y + 10 = 0. …$$

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9. TG EAPCET 2025Set eng-2025-05-02-FN1 markMCQ
   Q.If the normal drawn at P(8,16) to the parabola $y^2 = 32x$ meets the parabola again at Q, then the equation of the tangent drawn at Q to the parabola is (A) $x + 3y + 72 = 0$ (B) $x - y - 120 = 0$ (C) $3x - y - 264 = 0$ (D) $x + y - 24 = 0$

   ›Reveal solutionSolution

   The normal at $P$ meets the parabola again at $Q=(72,-48)$, and the tangent there is $x+3y+72=0$ — option (A).

   Setup. For $y^2=32x=4ax$ we have $a=8$, with parametric point $(8t^2,\,16t)$.

   Parameter of $P$. $P(8,16)$ gives $16t=16\Rightarrow t_1=1$.

   Second intersection. If the normal at $t_1$ meets the parabola again at $t_2$, then $t_2=-t_1-\dfrac{2}{t_1}=-1-2=-3$, so

   $Q=\big(8(-3)^2,\;16(-3)\big)=(72,-48).$

   Tangent at $Q$. The tangent at parameter $t$ is $ty=x+at^2$. With $t=-3,\ a=8$: …

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10. TG EAPCET 2025Set eng-2025-05-02-AN1 markMCQ
    Q.The four points whose position vectors are given by $2\vec{a} + 3\vec{b} - \vec{c}$, $\vec{a} - 2\vec{b} + 3\vec{c}$, $3\vec{a} + 4\vec{b} - 2\vec{c}$ and $\vec{a} - 6\vec{b} + 6\vec{c}$ are (A) Collinear (B) Coplanar (C) Vertices of a square (D) Vertices of a rectangle

    ›Reveal solutionSolution

    The four points are coplanar because the vectors connecting them are linearly dependent, meaning they all lie in the same plane. The correct option is (B).

    Concept and Intuition

    We are given four points defined by position vectors in terms of three independent vectors $\vec{a}, \vec{b}, \vec{c}$. The key question: are these points collinear, coplanar, or forming a special quadrilateral?

    Collinearity would mean all points lie on a single line — that’s very restrictive. Coplanarity means they all lie in some plane — a much weaker condition. Since we have three basis vectors, any point is in 3D space. Four points in 3D are always coplanar if the vectors from one point to the other three are linearly dependent (i.e., one is a combination of the other two).

    We can test this by picking one point as a reference and checking if the three difference vectors are linearly dependent. If they are, the points are coplanar. If they aren’t, the points are not coplanar (they form a tetrahedron).

    Let’s do exactly that.

    ---

    Step-by-step solution

    1. Label the points

    Let

    $P_1 = 2\vec{a} + 3\vec{b} - \vec{c}, \quad P_2 = \vec{a} - 2\vec{b} + 3\vec{c}, \quad P_3 = 3\vec{a} + 4\vec{b} - 2\vec{c}, \quad P_4 = \vec{a} - 6\vec{b} + 6\vec{c}.$

    2. Choose a reference point

    Take $P_1$ as the reference. Compute the vectors from $P_1$ to the other three points:

    $\vec{v}_2 = P_2 - P_1 = (\vec{a} - 2\vec{b} + 3\vec{c}) - (2\vec{a} + 3\vec{b} - \vec{c}) = -\vec{a} - 5\vec{b} + 4\vec{c}.$

    $\vec{v}_3 = P_3 - P_1 = (3\vec{a} + 4\vec{b} - 2\vec{c}) - (2\vec{a} + 3\vec{b} - \vec{c}) = \vec{a} + \vec{b} - \vec{c}.$

    $\vec{v}_4 = P_4 - P_1 = (\vec{a} - 6\vec{b} + 6\vec{c}) - (2\vec{a} + 3\vec{b} - \vec{c}) = -\vec{a} - 9\vec{b} + 7\vec{c}.$

    3. Check linear dependence

    We ask: can $\vec{v}_4$ be written as a combination of $\vec{v}_2$ and $\vec{v}_3$? That is, do there exist scalars $\alpha, \beta$ such that

    $\vec{v}_4 = \alpha \vec{v}_2 + \beta \vec{v}_3 ?$

    Substitute:

    $-\vec{a} - 9\vec{b} + 7\vec{c} = \alpha(-\vec{a} - 5\vec{b} + 4\vec{c}) + \beta(\vec{a} + \vec{b} - \vec{c}).$

    4. Equate coefficients

    Since $\vec{a}, \vec{b}, \vec{c}$ are independent, we equate coefficients:

    • For $\vec{a}$: $-1 = -\alpha + \beta$
    • For $\vec{b}$: $-9 = -5\alpha + \beta$
    • For $\vec{c}$: $7 = 4\alpha - \beta$

    5. Solve the system

    From the first equation: $\beta = \alpha - 1$.

    Substitute into the second:

    $-9 = -5\alpha + (\alpha - 1) \Rightarrow -9 = -4\alpha - 1 \Rightarrow -8 = -4\alpha \Rightarrow \alpha = 2$.

    Then $\beta = 2 - 1 = 1$.

    Check the third equation: $4(2) - 1 = 8 - 1 = 7$, which matches.

    So $\vec{v}_4 = 2\vec{v}_2 + \vec{v}_3$. The three vectors are linearly dependent. …

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11. TG EAPCET 2025Set eng-2025-05-02-AN1 markMCQ
    Q.The centres of all circles passing through the points of intersection of the circles $x^2 + y^2 + 2x - 2y + 1 = 0$ and $x^2 + y^2 - 2x + 2y - 2 = 0$ and having radius $\sqrt{14}$ lie on the curve (A) $x + y = 0$ (B) $y^2 = 4x - 2$ (C) $3x^2 + 5x = y$ (D) $2x^2 + 3y^2 = 7$

    ›Reveal solutionSolution

    Every circle through the two intersection points has its centre on the line joining the centres of the given circles, namely $x+y=0$ — option (A).

    The two circles.

    $C_1:\;x^2+y^2+2x-2y+1=0,\quad\text{centre }(-1,1),$

    $C_2:\;x^2+y^2-2x+2y-2=0,\quad\text{centre }(1,-1).$

    Family through the intersection. Any such circle is $C_1+\lambda C_2=0$, whose centre is the weighted average of the two centres:

    $h=\frac{-1+\lambda}{1+\lambda},\qquad k=\frac{1-\lambda}{1+\lambda}=-h.$

    Hence every centre satisfies $h+k=0$, i.e. the centres lie on the line

    $x+y=0.$ …

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12. TG EAPCET 2025Set eng-2025-05-02-AN1 markMCQ
    Q.For the parabola $y = x^2 - 3x + 2$, match the items in list-1 to that of the items in list-2. S is a focus, Z is intersection of axis and directrix, P is one end point of latus rectum, Q is the point on the parabola at which tangent is parallel to X-axis (A) A – I, B – II, C – III, D – IV (B) A – I, B – II, C – V, D – IV (C) A – II, B – V, C – III, D – IV (D) A – IV, B – V, C – III, D – I

    ›Reveal solutionSolution

    Writing the parabola in standard form gives focus $S=\left(\tfrac32,0\right)$, $Z=\left(\tfrac32,-\tfrac12\right)$, latus-rectum end $P=(2,0)$ and $Q=\left(\tfrac32,-\tfrac14\right)$; matching these to List-2 gives option (A).

    Standard form. Completing the square,

    $y=x^2-3x+2=\left(x-\tfrac32\right)^2-\tfrac14\quad\Rightarrow\quad\left(x-\tfrac32\right)^2=1\cdot\left(y+\tfrac14\right).$

    Comparing with $(x-h)^2=4a(y-k)$: vertex $\left(\tfrac32,-\tfrac14\right)$, $4a=1$ so $a=\tfrac14$; the parabola opens upward with axis $x=\tfrac32$.

    The four points.

    • Focus $S=(h,\,k+a)=\left(\tfrac32,\,-\tfrac14+\tfrac14\right)=\left(\tfrac32,\,0\right).$
    • Directrix $y=k-a=-\tfrac12$, so $Z=$ (axis $\cap$ directrix) $=\left(\tfrac32,\,-\tfrac12\right).$ …

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