Q.The centroid of a triangle $ABC$ is at the point $(1, 1, 1)$. If the coordinates of $A$ and $B$ are $(3, -5, 7)$ and $(-1, 7, -6)$, respectively, find the coordinates of the point $C$.

Showing the 12 most recent of 38 on this concept.

1. TG EAPCET 2026Set eng-2026-05-09-AN1 markMCQ
   Q.The external centre of similitude for the circles $x^2 + y^2 + 10x - 16y - 11 = 0$ and $x^2 + y^2 - 2x + 4y - 4 = 0$ is (A) $\left(\dfrac{5}{7}, -\dfrac{4}{7}\right)$ (B) $(-2, 3)$ (C) $\left(\dfrac{25}{7}, -\dfrac{44}{7}\right)$ (D) $(-3, 5)$

   ›Reveal solutionSolution

   The external centre of similitude lies on the line joining the centres and divides it externally in the ratio of the radii. For these two circles, the external centre is $\left(\frac{25}{7}, -\frac{44}{7}\right)$, which is option (C).

   The idea of a centre of similitude (also called a homothety centre) comes from scaling one circle into the other. For two circles, there are two such points: the internal centre (where the circles are scaled toward each other, dividing the line of centres internally in the ratio of radii) and the external centre (where one circle is scaled away from the other, dividing the line of centres externally in the same ratio). The external centre is the point from which the two circles appear to be scaled versions of each other, with the same orientation.

   To find it, we first need the centres and radii of both circles. Then we apply the section formula for external division.

   1. Rewrite each circle in standard form by completing the square.

      For the first circle:

      $x^2 + y^2 + 10x - 16y - 11 = 0$

      Group $x$ and $y$ terms:

      $(x^2 + 10x) + (y^2 - 16y) = 11$

      Complete the square:

      $(x^2 + 10x + 25) + (y^2 - 16y + 64) = 11 + 25 + 64$

      $(x + 5)^2 + (y - 8)^2 = 100$

      So centre $C_1 = (-5, 8)$ and radius $r_1 = \sqrt{100} = 10$.

      For the second circle:

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

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

      $(x^2 - 2x + 1) + (y^2 + 4y + 4) = 4 + 1 + 4$

      $(x - 1)^2 + (y + 2)^2 = 9$

      So centre $C_2 = (1, -2)$ and radius $r_2 = \sqrt{9} = 3$.

   2. The external centre of similitude divides the line segment $C_1C_2$ externally in the ratio $r_1 : r_2 = 10 : 3$.

      That means if the external centre is $P$, then $P$ lies on the line through $C_1$ and $C_2$ such that $PC_1 : PC_2 = 10 : 3$, but with $P$ outside the segment $C_1C_2$ on the side of the smaller circle.

      The formula for external division: if a point $P$ divides $A(x_1, y_1)$ and $B(x_2, y_2)$ externally in the ratio $m : n$, then

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

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2. TG EAPCET 2026Set eng-2026-05-10-AN1 markMCQ
   Q.If $A = (1,2,-3)$, $B = (2,3,-1)$, $C = (3,1,-2)$ are the vertices of a triangle ABC, then the area of triangle ABC is (A) $\dfrac{3\sqrt{3}}{2}$ (B) $\dfrac{2\sqrt{3}}{5}$ (C) $\dfrac{2\sqrt{5}}{3}$ (D) $\dfrac{3\sqrt{2}}{4}$

   ›Reveal solutionSolution

   The area of a triangle in 3D is half the magnitude of the cross product of two side vectors.

   For vertices $A(1,2,-3), B(2,3,-1), C(3,1,-2)$, the area is $\frac{3\sqrt{3}}{2}$, which corresponds to option (A).

   ---

   Concept and Intuition

   In 3D space, we can’t just use the base-times-height formula directly because the triangle is tilted. Instead, we use vectors: the area of a triangle formed by points $A, B, C$ is half the area of the parallelogram spanned by two side vectors, say $\overrightarrow{AB}$ and $\overrightarrow{AC}$. The area of that parallelogram is the magnitude of the cross product $|\overrightarrow{AB} \times \overrightarrow{AC}|$. So the triangle’s area is simply:

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

   This works because the cross product’s magnitude equals the product of the lengths of the two vectors times the sine of the angle between them — exactly the parallelogram area formula.

   ---

   Step-by-step solution

   1. Find the side vectors

   $\overrightarrow{AB} = B - A = (2-1,\; 3-2,\; -1 - (-3)) = (1, 1, 2)$

   $\overrightarrow{AC} = C - A = (3-1,\; 1-2,\; -2 - (-3)) = (2, -1, 1)$

   2. Compute the cross product The cross product $\overrightarrow{AB} \times \overrightarrow{AC}$ is given by the determinant:

   $\overrightarrow{AB} \times \overrightarrow{AC} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 1 & 1 & 2 \\ 2 & -1 & 1 \end{vmatrix}$

   Expand:

   • $\mathbf{i}$-component: $(1)(1) - (2)(-1) = 1 + 2 = 3$
   • $\mathbf{j}$-component: $-[(1)(1) - (2)(2)] = -[1 - 4] = -(-3) = 3$ (Careful: the $\mathbf{j}$ term has a minus sign in the determinant expansion.) …

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3. TG EAPCET 2026Set eng-2026-05-10-FN1 markMCQ
   Q.Let $A = (2,0,3)$, $B = (0,1,4)$ and $C = (5,6,0)$ be three points. If $L_1$ and $L_2$ are the lines bisecting the angles between $AB$ and $AC$, then the direction ratios of a line perpendicular to both $L_1$ and $L_2$ are (A) $(3,1,2)$ (B) $(1,5,2)$ (C) $(3,1,5)$ (D) $(1,3,5)$

   ›Reveal solutionSolution

   Both angle bisectors lie in the plane of $AB$ and $AC$, so a line perpendicular to both is normal to that plane: $\vec{AB}\times\vec{AC}\propto(3,1,5)$.

   $\vec{AB}=B-A=(-2,1,1),\qquad \vec{AC}=C-A=(3,6,-3).$

   The bisectors $L_1,L_2$ of the angle between $AB$ and $AC$ both lie in the plane spanned by $\vec{AB}$ and $\vec{AC}$. A line perpendicular to both bisectors must therefore be perpendicular to that whole plane, i.e. parallel to the normal $\vec{AB}\times\vec{AC}$: …

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4. TG EAPCET 2026Set eng-2026-05-11-FN1 markMCQ
   Q.Let ABC be an isosceles triangle. If $A = (2,3)$, $B = (3,2)$ and BC is its base then the locus of the point C is (A) a circle with radius 2 (B) a circle not containing the point $(1,4)$ (C) a parabola with vertex at $(2,3)$ (D) a parabola with focus at $(2,3)$

   ›Reveal solutionSolution

   In an isosceles triangle with base BC, the vertex A is equidistant from B and C. Using the distance formula, the condition $AB = AC$ gives the locus of C as a circle centered at A, which leads to the correct option.

   The key idea is that in an isosceles triangle, the two equal sides meet at the vertex. Here, A is given as the vertex, and BC is the base. That means sides AB and AC are equal. So point C must be such that its distance from A equals the fixed distance AB. That is the definition of a circle: all points at a constant distance from a fixed center. The center is A, and the radius is AB.

   Let’s work it out step by step.

   1. Find the fixed distance AB. $A = (2,3)$, $B = (3,2)$. Using the distance formula:

   $AB = \sqrt{(3-2)^2 + (2-3)^2} = \sqrt{1^2 + (-1)^2} = \sqrt{2}.$

   2. Set up the condition for C. Let $C = (x, y)$. Since $AB = AC$:

   $AC = \sqrt{(x-2)^2 + (y-3)^2} = \sqrt{2}.$

   3. Square both sides to get the equation of the locus.

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

   This is a circle with center at $(2,3)$ and radius $\sqrt{2}$.

   4. Check the options.
      • (A) says radius 2 — false, radius is $\sqrt{2}$.
      • (B) says a circle not containing $(1,4)$. Let’s test: put $(1,4)$ into the equation: $(1-2)^2 + (4-3)^2 = 1 + 1 = 2$, so $(1,4)$ lies on the circle. The option says “not containing”, which is false.
      • (C) and (D) mention a parabola — false, the locus is a circle. …

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5. TG EAPCET 2026Set eng-2026-05-11-FN1 markMCQ
   Q.Let $A(1,1)$ and $B(-1,-1)$ be the points of contact of the tangents drawn from a point $P$ to the circle $x^2 + y^2 - 2x + 2y - 2 = 0$. If $C$ is the centre of the circle, then the centre of the circle passing through the points $A$, $B$, $C$ and $P$ is (A) $\left( -\dfrac{1}{6}, -\dfrac{1}{3} \right)$ (B) $(0,0)$ (C) $\left( -\dfrac{4}{3}, -\dfrac{1}{3} \right)$ (D) $\left( \dfrac{1}{6}, \dfrac{4}{3} \right)$

   ›Reveal solutionSolution

   The circle through $A$, $B$, $C$, $P$ has $PC$ as diameter, so its centre is the midpoint of $PC$, namely $(0,0)$.

   The circle is $x^2+y^2-2x+2y-2=0$, with centre $C=(1,-1)$ and radius $r=\sqrt{1+1+2}=2$.

   Find $P$. $P$ is the intersection of the tangents at $A(1,1)$ and $B(-1,-1)$. Using $T=0$:

   Tangent at $A(1,1)$: $x(1)+y(1)-(x+1)+(y+1)-2=0 \Rightarrow 2y-2=0 \Rightarrow y=1.$

   Tangent at $B(-1,-1)$: $x(-1)+y(-1)-(x-1)+(y-1)-2=0 \Rightarrow -2x-2=0 \Rightarrow x=-1.$

   So $P=(-1,1)$. …

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6. TG EAPCET 2026Set eng-2026-05-11-AN1 markMCQ
   Q.If P$(x, y, z)$ is the intersection of the lines $\vec{r} = (2\vec{i} - 2\vec{j} + 3\vec{k}) + t(\vec{i} - 3\vec{j} + \vec{k})$ and $\vec{r} = (2\vec{j} - \vec{k}) + s(\vec{i} - \vec{j} + 3\vec{k})$, then $x + y + z =$ (A) $2$ (B) $5$ (C) $3$ (D) $4$

   ›Reveal solutionSolution

   The lines meet at $(1,1,2)$, so $x+y+z=4$ (D).

   Points on the two lines are

   $L_1:\ (2+t,\ -2-3t,\ 3+t),\qquad L_2:\ (s,\ 2-s,\ -1+3s).$

   Equating coordinates:

   $2+t=s,\qquad -2-3t=2-s,\qquad 3+t=-1+3s.$

   From the first, $s=2+t$. Substituting into the second:

   $-2-3t=2-(2+t)=-t\ \Rightarrow\ -2=2t\ \Rightarrow\ t=-1,\quad s=1.$ …

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7. TG EAPCET 2026Set eng-2026-05-11-AN1 markMCQ
   Q.If the locus of the midpoints of the chords of the circle $S \equiv x^2 + y^2 - 6x - 8y - 11 = 0$ which subtend a right angle at $A(1,2)$ is another circle $S' = 0$, then the centre of $S' = 0$ (A) is the midpoint of the line segment joining A and the centre of $S = 0$ (B) Divides the line segment joining A and the centre of $S = 0$ in the ratio $1:2$ (C) Lies outside the circle $S = 0$ (D) A vertex of the triangle having A and the centre of $S = 0$ as other two vertices

   ›Reveal solutionSolution

   The locus of midpoints of chords of a circle that subtend a right angle at a fixed point is a circle whose centre is the midpoint of the fixed point and the original circle’s centre. The correct option is (A).

   We are given the circle

   $S: x^2 + y^2 - 6x - 8y - 11 = 0.$

   Rewrite it in centre-radius form:

   $(x - 3)^2 + (y - 4)^2 = 36,$

   so centre $O = (3,4)$ and radius $R = 6$.

   We consider chords of this circle that subtend a right angle at a fixed point $A(1,2)$. The question asks for the locus of the midpoints of all such chords, and then to identify the centre of that locus.

   ---

   Concept and intuition

   For a chord of a circle, the midpoint is the foot of the perpendicular from the centre to the chord. If a chord subtends a right angle at a point $A$, then the chord is seen from $A$ under $90^\circ$. A powerful geometric fact: the locus of points $M$ (midpoints of chords) such that the chord subtends a right angle at a fixed point $A$ is a circle whose diameter is the segment joining $A$ to the centre $O$ of the original circle. This is because the condition translates into a fixed power relationship or a right-angle condition in a triangle formed by $A$, $O$, and $M$.

   Let’s derive it step by step.

   ---

   Step-by-step derivation

   1. Set up coordinates and variables

      Let $M(h,k)$ be the midpoint of a chord of circle $S$. The chord is perpendicular to $OM$ (since the line from centre to midpoint of a chord is perpendicular to the chord). So the chord’s direction is perpendicular to vector $\overrightarrow{OM} = (h-3, k-4)$.

   2. Equation of the chord with given midpoint

      The chord with midpoint $M$ has equation (using the chord-midpoint formula for a circle):

   $T = S_1,$

   where $T$ is the equation of the tangent-like form at $M$. For circle $x^2+y^2-6x-8y-11=0$, the chord with midpoint $(h,k)$ is:

   $xh + yk - 3(x+h) - 4(y+k) - 11 = h^2 + k^2 - 6h - 8k - 11.$

   Simplify:

   $(h-3)x + (k-4)y - (3h+4k+11) = h^2 + k^2 - 6h - 8k - 11.$

   This line passes through the endpoints of the chord.

   3. Condition that chord subtends a right angle at $A(1,2)$

      The chord’s endpoints $P$ and $Q$ satisfy $\angle PAQ = 90^\circ$. This is equivalent to the pair of lines $AP$ and $AQ$ being perpendicular. A standard method: the combined equation of lines $AP$ and $AQ$ is obtained by homogenizing the circle equation with the chord as the line pair through $A$. But a simpler approach uses the property of the circle with diameter $PQ$: if $\angle PAQ = 90^\circ$, then $A$ lies on the circle with diameter $PQ$. That circle’s centre is $M$ and radius is $MP$. So $AM = MP$.

   4. Express $MP$ in terms of $M$ and $O$

      Since $P$ lies on the original circle, and $M$ is the midpoint of chord $PQ$, we have $OM \perp PQ$. In right triangle $OMP$,

   $OP^2 = OM^2 + MP^2.$

   But $OP = R = 6$, so

   $MP^2 = 36 - OM^2.$

   5. Apply the right-angle condition at $A$ From step 3, $AM = MP$. So

   $AM^2 = MP^2 = 36 - OM^2.$

   Compute:

   $AM^2 = (h-1)^2 + (k-2)^2,$

   $OM^2 = (h-3)^2 + (k-4)^2.$

   Hence:

   $(h-1)^2 + (k-2)^2 = 36 - \left[(h-3)^2 + (k-4)^2\right].$

   6. Simplify to find the locus Expand: $$(h^2 - 2h + 1) + (k^2 - 4k + 4) = 36 - \left[(h^2 - 6h + 9) + (k^2 - 8k + 16)\right]. …$$

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8. TG EAPCET 2025Set eng-2025-05-02-FN1 markMCQ
   Q.If the equation of the circle passing through the points $(-1,0),\,(-1,1),\,(1,1)$ is $ax^2+ay^2+2gx+2fy-2=0$ then $a=$ (A) $1$ (B) $-1$ (C) $2$ (D) $-2$

   ›Reveal solutionSolution

   The key idea is to substitute the three given points into the general circle equation, solve for the unknown parameters, and find that $a = 1$. The correct option is (A).

   We are told the circle has equation

   $ax^2 + ay^2 + 2gx + 2fy - 2 = 0.$

   This is a general second-degree equation representing a circle (coefficients of $x^2$ and $y^2$ are equal, and no $xy$ term). Our job: find $a$ such that the three given points lie on this circle.

   Concept & Intuition

   A circle is uniquely determined by three non-collinear points. Plugging each point into the equation gives a linear equation in the unknowns $a, g, f$. Since we only need $a$, we can eliminate $g$ and $f$ by subtracting equations or solving the system. The constant term $-2$ is already fixed, so the three points will force a specific $a$.

   ---

   Step-by-step solution

   1. Substitute $(-1,0)$

   $a(-1)^2 + a(0)^2 + 2g(-1) + 2f(0) - 2 = 0$

   $a - 2g - 2 = 0 \quad\Rightarrow\quad a - 2g = 2. \qquad(1)$

   2. Substitute $(-1,1)$

   $a(-1)^2 + a(1)^2 + 2g(-1) + 2f(1) - 2 = 0$

   $a + a - 2g + 2f - 2 = 0$

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

   3. Substitute $(1,1)$

   $a(1)^2 + a(1)^2 + 2g(1) + 2f(1) - 2 = 0$

   $a + a + 2g + 2f - 2 = 0$

   $2a + 2g + 2f = 2 \quad\Rightarrow\quad 2a + 2g + 2f = 2. \qquad(3)$

   4. Solve the system From (1): $a - 2g = 2$ → $2g = a - 2$. Subtract (2) from (3) to eliminate $2f$:

   $(2a + 2g + 2f) - (2a - 2g + 2f) = 2 - 2$

   $4g = 0 \quad\Rightarrow\quad g = 0.$

   Then from $2g = a - 2$ we get $0 = a - 2$ → $a = 2$. Wait—this seems to give $a = 2$. But let’s check consistency with the other equations.

   With $g = 0$ and $a = 2$, equation (1) becomes $2 - 0 = 2$ (okay).

   Equation (2): $2(2) - 0 + 2f = 2$ → $4 + 2f = 2$ → $2f = -2$ → $f = -1$.

   Equation (3): $2(2) + 0 + 2(-1) = 4 - 2 = 2$ (okay). …

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9. TG EAPCET 2025Set eng-2025-05-02-AN1 markMCQ
   Q.If L is a line common to the planes $3x + 4y + 7z = 1$, $x - y + z = 5$ then the direction ratios of the line L are (A) $(16, 0, -1)$ (B) $(11, 4, -7)$ (C) $(2, 5, 1)$ (D) $(4, -7, 11)$

   ›Reveal solutionSolution

   The line common to two planes is their line of intersection, which is perpendicular to both normals. Taking the cross product of the normals gives direction ratios $(11, 4, -7)$, matching option (B).

   The key idea: A line that lies in both planes must be perpendicular to the normal vectors of both planes. Therefore, its direction vector is parallel to the cross product of the two normals. This is a standard and reliable method — no need to solve for points or parametrics unless you want extra work.

   ---

   1. Identify the normal vectors

      For the plane $3x + 4y + 7z = 1$, the normal vector is $\vec{n}_1 = (3, 4, 7)$.

      For the plane $x - y + z = 5$, the normal vector is $\vec{n}_2 = (1, -1, 1)$.

   2. Find a direction vector for the line of intersection

      The line common to both planes is perpendicular to both normals, so its direction vector $\vec{d}$ is parallel to $\vec{n}_1 \times \vec{n}_2$.

      Compute the cross product:

   $\vec{d} = \vec{n}_1 \times \vec{n}_2 = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 3 & 4 & 7 \\ 1 & -1 & 1 \end{vmatrix}$

   $= \mathbf{i}(4 \cdot 1 - 7 \cdot (-1)) - \mathbf{j}(3 \cdot 1 - 7 \cdot 1) + \mathbf{k}(3 \cdot (-1) - 4 \cdot 1)$

   $= \mathbf{i}(4 + 7) - \mathbf{j}(3 - 7) + \mathbf{k}(-3 - 4)$

   $$$$

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10. TG EAPCET 2025Set eng-2025-05-03-AN1 markMCQ
    Q.In a triangle ABC, if $\overrightarrow{BC} = \hat{i} - 2\hat{j} + 2\hat{k}$ and $\overrightarrow{CA} = 6\hat{i} + 3\hat{j} - 2\hat{k}$, then the perimeter of the triangle is (A) $5(2+\sqrt{3})$ (B) $5(2+\sqrt{2})$ (C) $\sqrt{10}(3+\sqrt{10})$ (D) $10(2+\sqrt{5})$

    ›Reveal solutionSolution

    The perimeter is the sum of the lengths of all three sides. We are given two side vectors; the third side vector is found by the triangle law of vector addition. After computing lengths, the perimeter simplifies to $5(2+\sqrt{2})$, which matches option (B).

    We are given vectors for two sides of triangle $ABC$:

    $\overrightarrow{BC} = \hat{i} - 2\hat{j} + 2\hat{k}, \quad \overrightarrow{CA} = 6\hat{i} + 3\hat{j} - 2\hat{k}.$

    The perimeter is the sum of the lengths of sides $AB$, $BC$, and $CA$. We already have $BC$ and $CA$; we need $AB$.

    Key idea: In a triangle, the vectors between vertices satisfy

    $\overrightarrow{AB} + \overrightarrow{BC} + \overrightarrow{CA} = \mathbf{0}.$

    This is because going from $A$ to $B$ to $C$ and back to $A$ brings you to the starting point. So we can solve for $\overrightarrow{AB}$.

    1. Find $\overrightarrow{AB}$. From $\overrightarrow{AB} + \overrightarrow{BC} + \overrightarrow{CA} = \mathbf{0}$, we have

    $\overrightarrow{AB} = -\overrightarrow{BC} - \overrightarrow{CA}.$

    Substitute the given vectors:

    $\overrightarrow{AB} = -(\hat{i} - 2\hat{j} + 2\hat{k}) - (6\hat{i} + 3\hat{j} - 2\hat{k}).$

    Simplify component-wise:

    • $\hat{i}$: $-1 - 6 = -7$
    • $\hat{j}$: $2 - 3 = -1$
    • $\hat{k}$: $-2 + 2 = 0$ So

    $\overrightarrow{AB} = -7\hat{i} - \hat{j} + 0\hat{k}.$

    2. Compute the length of each side. The length (magnitude) of a vector $a\hat{i}+b\hat{j}+c\hat{k}$ is $\sqrt{a^2+b^2+c^2}$.
       • For $BC$:

    $|\overrightarrow{BC}| = \sqrt{1^2 + (-2)^2 + 2^2} = \sqrt{1+4+4} = \sqrt{9} = 3.$

    • For $CA$:

    $|\overrightarrow{CA}| = \sqrt{6^2 + 3^2 + (-2)^2} = \sqrt{36+9+4} = \sqrt{49} = 7.$

    • For $AB$: $$$$

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11. TG EAPCET 2025Set eng-2025-05-03-FN1 markMCQ
    Q.In a triangle ABC, if $\overrightarrow{BC} = \hat{i} - 2\hat{j} + 2\hat{k}$ and $\overrightarrow{CA} = 6\hat{i} + 3\hat{j} - 2\hat{k}$, then the perimeter of the triangle is (A) $\sqrt{10}\left(3 + \sqrt{10}\right)$ (B) $5\left(2 + \sqrt{3}\right)$ (C) $5\left(2 + \sqrt{2}\right)$ (D) $10\left(2 + \sqrt{5}\right)$

    ›Reveal solutionSolution

    The perimeter is the sum of the lengths of the three sides. Given two side vectors, the third is found by vector addition; then compute each length and sum them. The result matches option (C).

    We are given two side vectors of triangle ABC:

    $\overrightarrow{BC} = \hat{i} - 2\hat{j} + 2\hat{k}, \quad \overrightarrow{CA} = 6\hat{i} + 3\hat{j} - 2\hat{k}.$

    The perimeter is the sum of the lengths of sides $AB$, $BC$, and $CA$. We already have two vectors; we need the third, $\overrightarrow{AB}$.

    ---

    1. Find the third side vector $\overrightarrow{AB}$

    In any triangle, the vectors around the triangle sum to zero:

    $\overrightarrow{AB} + \overrightarrow{BC} + \overrightarrow{CA} = \mathbf{0}.$

    So

    $\overrightarrow{AB} = -\overrightarrow{BC} - \overrightarrow{CA}.$

    Substitute:

    $\overrightarrow{AB} = -(\hat{i} - 2\hat{j} + 2\hat{k}) - (6\hat{i} + 3\hat{j} - 2\hat{k}) = -\hat{i} + 2\hat{j} - 2\hat{k} - 6\hat{i} - 3\hat{j} + 2\hat{k}.$

    Combine components:

    $\overrightarrow{AB} = (-1-6)\hat{i} + (2-3)\hat{j} + (-2+2)\hat{k} = -7\hat{i} - \hat{j} + 0\hat{k}.$

    Thus

    $\overrightarrow{AB} = -7\hat{i} - \hat{j}.$

    ---

    2. Compute the lengths of each side

    Length of a vector $\vec{v} = a\hat{i} + b\hat{j} + c\hat{k}$ is $\sqrt{a^2 + b^2 + c^2}$.

    • For $BC$:

    $|\overrightarrow{BC}| = \sqrt{1^2 + (-2)^2 + 2^2} = \sqrt{1 + 4 + 4} = \sqrt{9} = 3.$

    • For $CA$:

    $|\overrightarrow{CA}| = \sqrt{6^2 + 3^2 + (-2)^2} = \sqrt{36 + 9 + 4} = \sqrt{49} = 7.$

    • For $AB$:

    $$$$

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12. TG EAPCET 2025Set eng-2025-05-03-FN1 markMCQ
    Q.If $(\alpha, \beta)$ is the centre of the circle which passes through the point $(1, -1)$ and cuts the circles $x^2 + y^2 + 2x - 3y - 5 = 0$, $x^2 + y^2 - 3x + 2y + 1 = 0$ orthogonally, then $\alpha - 5\beta =$ (A) $-10$ (B) $10$ (C) $-11$ (D) $5$

    ›Reveal solutionSolution

    The centre $(\alpha,\beta)$ lies on the radical axis of the two given circles (since it cuts both orthogonally), and also satisfies the condition that the power of the centre w.r.t. each circle equals the square of the radius of the orthogonal circle. Solving gives $\alpha - 5\beta = -11$, so the answer is (C).

    Concept & Intuition

    Two circles cut orthogonally if at their intersection points the tangents are perpendicular. The algebraic condition is:

    $2g_1g_2 + 2f_1f_2 = c_1 + c_2$ when circles are $x^2+y^2+2gx+2fy+c=0$.

    If a circle with centre $(\alpha,\beta)$ and radius $r$ cuts a given circle orthogonally, then the square of the distance between centres equals $r^2$ + (radius of given circle)$^2$. Equivalently, the power of $(\alpha,\beta)$ w.r.t. the given circle equals $r^2$. Since $r^2$ is the same for both given circles (it’s the radius of the same orthogonal circle), the powers of $(\alpha,\beta)$ w.r.t. the two circles must be equal. That equality is precisely the radical axis of the two circles. So $(\alpha,\beta)$ lies on that line. Then we use the orthogonal condition with one circle to find the exact centre.

    Step-by-step solution

    1. Write the circles in standard form

       Circle 1: $x^2+y^2+2x-3y-5=0$

       Here $2g_1=2 \Rightarrow g_1=1$, $2f_1=-3 \Rightarrow f_1=-\frac{3}{2}$, $c_1=-5$.

       Centre $C_1=(-1,\frac{3}{2})$, radius $R_1=\sqrt{g_1^2+f_1^2-c_1}=\sqrt{1+\frac{9}{4}+5}=\sqrt{\frac{4+9+20}{4}}=\sqrt{\frac{33}{4}}=\frac{\sqrt{33}}{2}$.

       Circle 2: $x^2+y^2-3x+2y+1=0$

       Here $2g_2=-3 \Rightarrow g_2=-\frac{3}{2}$, $2f_2=2 \Rightarrow f_2=1$, $c_2=1$.

       Centre $C_2=(\frac{3}{2},-1)$, radius $R_2=\sqrt{\frac{9}{4}+1-1}=\sqrt{\frac{9}{4}}=\frac{3}{2}$.

    2. Orthogonal condition for a circle with centre $(\alpha,\beta)$ and radius $r$

       For circle 1: $(\alpha+1)^2+(\beta-\frac{3}{2})^2 = r^2 + R_1^2$

       $\Rightarrow (\alpha+1)^2+(\beta-\frac{3}{2})^2 = r^2 + \frac{33}{4}$. …(1)

       For circle 2: $(\alpha-\frac{3}{2})^2+(\beta+1)^2 = r^2 + R_2^2$

       $\Rightarrow (\alpha-\frac{3}{2})^2+(\beta+1)^2 = r^2 + \frac{9}{4}$. …(2)

    3. Eliminate $r^2$ by equating the left-hand sides minus the respective $R^2$

       From (1): $r^2 = (\alpha+1)^2+(\beta-\frac{3}{2})^2 - \frac{33}{4}$

       From (2): $r^2 = (\alpha-\frac{3}{2})^2+(\beta+1)^2 - \frac{9}{4}$

       Set equal:

       $(\alpha+1)^2+(\beta-\frac{3}{2})^2 - \frac{33}{4} = (\alpha-\frac{3}{2})^2+(\beta+1)^2 - \frac{9}{4}$

    4. Simplify to get the radical axis

       Expand:

       $\alpha^2+2\alpha+1 + \beta^2-3\beta+\frac{9}{4} - \frac{33}{4} = \alpha^2-3\alpha+\frac{9}{4} + \beta^2+2\beta+1 - \frac{9}{4}$

       Cancel $\alpha^2,\beta^2$ and constants:

       $2\alpha+1 -3\beta + \frac{9}{4}-\frac{33}{4} = -3\alpha + 2\beta + 1 + \frac{9}{4}-\frac{9}{4}$

       Simplify constants: $\frac{9}{4}-\frac{33}{4} = -\frac{24}{4} = -6$

       So LHS: $2\alpha+1-3\beta-6 = 2\alpha -3\beta -5$

       RHS: $-3\alpha+2\beta+1$

       Equation: $2\alpha -3\beta -5 = -3\alpha+2\beta+1$

       $\Rightarrow 5\alpha -5\beta = 6$

       $\Rightarrow \alpha - \beta = \frac{6}{5}$. …(3)

    5. Use the fact that the circle also passes through $(1,-1)$

       Distance from $(\alpha,\beta)$ to $(1,-1)$ equals $r$:

       $r^2 = (\alpha-1)^2+(\beta+1)^2$. …(4)

    6. Plug $r^2$ into orthogonal condition with one circle (say circle 1)

       From (1): $(\alpha+1)^2+(\beta-\frac{3}{2})^2 = (\alpha-1)^2+(\beta+1)^2 + \frac{33}{4}$

       Expand:

       $\alpha^2+2\alpha+1 + \beta^2-3\beta+\frac{9}{4} = \alpha^2-2\alpha+1 + \beta^2+2\beta+1 + \frac{33}{4}$ …

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