Q.(b) If $(x - a)^2 + (y - b)^2 = c^2$, for some $c > 0$, prove that $\dfrac{\left[1 + \left(\dfrac{dy}{dx}\right)^2\right]^{3/2}}{\dfrac{d^2y}{dx^2}}$ is a constant independent of $a$ and $b$.

Showing the 12 most recent of 18 on this concept.

1. TG EAPCET 2022Set eng-2022-07-18-FN1 markMCQ
   Q.A circle passes through the points (1, 2), (3, 4). If its centre lies on the line $x - y + 3 = 0$, then its radius is equal to (A) 4 (B) 3 (C) 1 (D) 2

   ›Reveal solutionSolution

   The centre of the circle lies on the given line and is equidistant from the two given points. Solving the resulting system gives the centre (1, 4), and the distance to either point yields the radius 2. The correct option is (D).

   ---

   The key idea is that a circle’s centre is equidistant from any two points on the circle. Here we know two points and a line the centre must lie on. So we have two conditions: the centre satisfies the line equation, and its distances to the two points are equal. That gives us two equations in the two coordinates of the centre.

   Let’s work through it step by step.

   1. Set up the centre coordinates Let the centre be $C(h, k)$. Since it lies on the line $x - y + 3 = 0$, we have

   $h - k + 3 = 0 \quad \Rightarrow \quad k = h + 3.$

   2. Use equal distances The distances from $C$ to $A(1,2)$ and $B(3,4)$ must be equal (both are the radius $r$). So

   $(h-1)^2 + (k-2)^2 = (h-3)^2 + (k-4)^2.$

   3. Substitute $k = h+3$

   $(h-1)^2 + ((h+3)-2)^2 = (h-3)^2 + ((h+3)-4)^2$

   Simplify the $k$-terms:

   $(h-1)^2 + (h+1)^2 = (h-3)^2 + (h-1)^2.$

   4. Cancel the common term Notice $(h-1)^2$ appears on both sides. Cancel it:

   $(h+1)^2 = (h-3)^2.$

   5. Solve for $h$ Taking square roots gives two possibilities:

   $h+1 = \pm (h-3).$

   • Case 1: $h+1 = h-3$ gives $1 = -3$, impossible.
   • Case 2: $h+1 = -(h-3)$ gives $h+1 = -h+3$ $\Rightarrow$ $2h = 2$ $\Rightarrow$ $h = 1$.

   So $h = 1$, and then $k = h+3 = 4$. The centre is $(1,4)$.

   6. Find the radius The radius is the distance from $(1,4)$ to either point, say $(1,2)$: $$$$

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2. TG EAPCET 2025Set eng-2025-05-04-FN1 markMCQ
   Q.If $P\left(\dfrac{7}{5}, \dfrac{6}{5}\right)$ is the inverse point of $A(1,2)$ with respect to a circle with centre $C(2,0)$, then the radius of that circle is (A) $\sqrt{3}$ (B) $1$ (C) $9$ (D) $3$

   ›Reveal solutionSolution

   The inverse of a point with respect to a circle satisfies $CP \times CA = r^2$. Using the given coordinates, the radius is found to be $3$.

   The idea of an inverse point is geometric: given a circle with centre $C$ and radius $r$, two points $A$ and $P$ are inverses if they lie on the same ray from $C$ and satisfy $CA \cdot CP = r^2$. This means that if you know the distances from the centre to both points, the radius is simply the geometric mean of those distances.

   Here, we are given $C(2,0)$, $A(1,2)$, and $P\left(\frac{7}{5}, \frac{6}{5}\right)$. The first step is to check that $P$ lies on the ray from $C$ through $A$ — but even if it does, the formula $CA \cdot CP = r^2$ holds for any inverse pair, so we can compute directly.

   1. Find $CA$: the distance from $C(2,0)$ to $A(1,2)$ is

   $CA = \sqrt{(1-2)^2 + (2-0)^2} = \sqrt{(-1)^2 + 2^2} = \sqrt{1+4} = \sqrt{5}.$

   2. Find $CP$: the distance from $C(2,0)$ to $P\left(\frac{7}{5}, \frac{6}{5}\right)$ is $$CP = \sqrt{\left(\frac{7}{5} - 2\right)^2 + \left(\frac{6}{5} - 0\right)^2} = \sqrt{\left(\frac{7}{5} - \frac{10}{5}\right)^2 + \left(\frac{6}{5}\right)^2} …$$

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3. TG EAPCET 2026Set eng-2026-05-09-FN1 markMCQ
   Q.Let P be any point on the circle $x^2 + y^2 = 16$ and $A = (1,2)$. If the locus of the point which divides the line segment AP in the ratio $3:2$ is a circle, then its radius is (A) $5$ (B) $4$ (C) $\dfrac{12}{5}$ (D) $\dfrac{13}{4}$

   ›Reveal solutionSolution

   The locus is a circle whose radius is the distance from the fixed point A to the moving point P scaled by the ratio of division; the answer is $\frac{12}{5}$, option (C).

   We have a fixed point $A = (1,2)$ and a variable point $P$ on the circle $x^2 + y^2 = 16$. A point $Q$ divides segment $AP$ in the ratio $3:2$. The question asks for the radius of the circle that $Q$ traces as $P$ moves.

   Concept & Intuition

   When a point divides a segment joining a fixed point and a moving point in a fixed ratio, the locus is a scaled and possibly shifted version of the original path. Here, $P$ moves on a circle centered at the origin with radius $4$. The division point $Q$ will also trace a circle, but its center will be somewhere between $A$ and the origin, and its radius will be a fraction of $4$. The key is to express $Q$ in terms of $P$ and then use the fact that $P$ satisfies $x^2+y^2=16$.

   ---

   1. Set up the section formula Let $P = (x_1, y_1)$ and $Q = (x, y)$. The ratio $AQ:QP = 3:2$ means $Q$ divides $AP$ internally in the ratio $3:2$ (from $A$ to $P$). Using the section formula:

   $Q = \left( \frac{2 \cdot A + 3 \cdot P}{3+2} \right)$

   because the weights are opposite to the segments: the coordinate of $Q$ is a weighted average of $A$ and $P$ with weights proportional to the opposite parts.

   So:

   $x = \frac{2(1) + 3x_1}{5}, \quad y = \frac{2(2) + 3y_1}{5}.$

   2. Solve for $P$ in terms of $Q$ Multiply both equations by 5:

   $5x = 2 + 3x_1 \quad\Rightarrow\quad 3x_1 = 5x - 2 \quad\Rightarrow\quad x_1 = \frac{5x - 2}{3}.$

   Similarly,

   $y_1 = \frac{5y - 4}{3}.$

   3. Use the fact that $P$ lies on the given circle Since $x_1^2 + y_1^2 = 16$, substitute:

   $\left( \frac{5x - 2}{3} \right)^2 + \left( \frac{5y - 4}{3} \right)^2 = 16.$

   Multiply through by 9:

   $(5x - 2)^2 + (5y - 4)^2 = 144.$

   4. Identify the locus This is the equation of a circle. Divide through by $5^2$ to see the radius more clearly, or complete the square directly. Expand:

   $25x^2 - 20x + 4 + 25y^2 - 40y + 16 = 144.$

   Simplify:

   $25x^2 + 25y^2 - 20x - 40y + 20 = 144.$

   $25x^2 + 25y^2 - 20x - 40y - 124 = 0.$

   Divide by 25:

   $x^2 + y^2 - \frac{4}{5}x - \frac{8}{5}y - \frac{124}{25} = 0.$

   5. Find the radius …

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4. TG EAPCET 2025Set eng-2025-05-04-AN1 markMCQ
   Q.If P$\left(\dfrac{7}{5}, \dfrac{6}{5}\right)$ is the inverse point of A(1,2) with respect to a circle with centre C(2, 0), then the radius of that circle is (A) 9 (B) 3 (C) $\sqrt{3}$ (D) 1

   ›Reveal solutionSolution

   The key idea is that for inverse points with respect to a circle, the product of distances from the centre to each point equals the square of the radius. Using the given coordinates, we find the radius is 3, so the correct option is (B).

   Concept & Intuition

   Inversion in a circle is a transformation where a point $P$ and its inverse $P'$ satisfy $CP \cdot CP' = r^2$, with $C$ the centre and $r$ the radius. Here, we are told $P$ is the inverse of $A$ with respect to a circle centred at $C$. So the distances from $C$ to $A$ and to $P$ multiply to $r^2$. We compute both distances, then solve for $r$.

   Step-by-step solution

   1. Identify the given points

      Centre: $C(2,0)$

      Point $A(1,2)$

      Inverse point $P\left(\frac{7}{5}, \frac{6}{5}\right)$

   2. Compute distance $CA$

   $CA = \sqrt{(1-2)^2 + (2-0)^2} = \sqrt{(-1)^2 + 2^2} = \sqrt{1+4} = \sqrt{5}$

   3. Compute distance $CP$

   $CP = \sqrt{\left(\frac{7}{5} - 2\right)^2 + \left(\frac{6}{5} - 0\right)^2}$

   First, $\frac{7}{5} - 2 = \frac{7}{5} - \frac{10}{5} = -\frac{3}{5}$

   Then, $\left(-\frac{3}{5}\right)^2 = \frac{9}{25}$

   Next, $\left(\frac{6}{5}\right)^2 = \frac{36}{25}$

   Sum: $\frac{9}{25} + \frac{36}{25} = \frac{45}{25} = \frac{9}{5}$ …

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5. TG EAPCET 2023Set eng-2023-05-12-FN1 markMCQ
   Q.If $C(\alpha, \beta)$ $(\alpha < 0)$ is the centre of the circle that touches the Y-axis at $(0,3)$ and makes an intercept of length $2$ units on positive X-axis, then $(\alpha, \beta) =$ (A) $(-3, \sqrt{10})$ (B) $(-3, -\sqrt{10})$ (C) $(-\sqrt{10}, 3)$ (D) $(-\sqrt{10}, -3)$

   ›Reveal solutionSolution

   The circle touches the Y‑axis at (0,3) and cuts a chord of length 2 on the positive X‑axis; using geometry of tangency and chord properties gives centre (−√10, 3), so option (C) is correct.

   ---

   Concept & Intuition

   A circle that touches the Y‑axis has its centre at a horizontal distance equal to its radius from that axis. The intercept on the X‑axis is a chord; the perpendicular from the centre to that chord bisects it. Combining these two geometric facts lets us solve for the centre coordinates without heavy algebra.

   ---

   Step‑by‑step reasoning

   1. Interpret “touches the Y‑axis at (0,3)” Touching means the Y‑axis is tangent to the circle at (0,3). The radius to the point of tangency is perpendicular to the tangent, so it is horizontal. Hence the centre must have the same y‑coordinate as the point of tangency:

   $\beta = 3.$

   The distance from the centre to the Y‑axis equals the radius $r$. Since the centre is $(\alpha,3)$ and the Y‑axis is $x=0$, we have

   $r = |\alpha|.$

   Given $\alpha < 0$, we have $r = -\alpha$ (positive).

   2. Interpret “intercept of length 2 on positive X‑axis” The circle cuts the X‑axis at two points; the segment between them has length 2. The X‑axis is a chord of the circle. The perpendicular from the centre to a chord bisects it. The centre is $(\alpha,3)$, so its perpendicular distance to the X‑axis is $|3| = 3$. Half the chord length is $1$. By the chord‑radius relation:

   $r^2 = (\text{distance from centre to chord})^2 + (\text{half‑chord})^2.$

   Hence

   $r^2 = 3^2 + 1^2 = 9 + 1 = 10.$

   3. Find $\alpha$ …

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6. TG EAPCET 2025Set eng-2025-05-02-FN1 markMCQ
   Q.If A, B are the points of contact of the tangents drawn from the point $(-3,1)$ to the circle $x^2 + y^2 - 4x + 2y - 4 = 0$, then the equation of the circumcircle of the triangle PAB is (A) $x^2 + y^2 - 6x + 2y - 6 = 0$ (B) $x^2 + y^2 - x + 7 = 0$ (C) $x^2 + y^2 + x - 7 = 0$ (D) $x^2 + y^2 + 6x - 2y - 6 = 0$

   ›Reveal solutionSolution

   The points of contact A and B, together with P and the centre C of the given circle, are concyclic on the circle whose diameter is CP (because a radius is perpendicular to a tangent). So the circumcircle of triangle PAB is the circle on CP as diameter, which is $x^2 + y^2 + x - 7 = 0$ — option (C).

   We are given the circle

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

   and an external point $P(-3,1)$. Tangents from $P$ touch the circle at $A$ and $B$, and we need the circumcircle of triangle $PAB$.

   Key idea. The radius to a point of contact is perpendicular to the tangent there, so $\angle CAP = \angle CBP = 90^\circ$, where $C$ is the centre. A right angle subtended at $A$ and $B$ by the segment $CP$ means $A$, $B$ (and $P$ itself) all lie on the circle having $CP$ as a diameter. Hence the circumcircle of $PAB$ is exactly the circle on diameter $CP$.

   Step 1 — Centre of the given circle.

   $(x-2)^2 + (y+1)^2 = 9 \quad\Rightarrow\quad C = (2,-1),\ r = 3.$

   Step 2 — Circle on diameter $CP$.

   With $C=(2,-1)$ and $P=(-3,1)$, the circle on diameter $CP$ is

   $$$$

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7. TG EAPCET 2022Set eng-2022-07-20-AN1 markMCQ
   Q.The line $4x + 3y - 4 = 0$ divides the circumference of a circle in the ratio $1 : 2$. If $C(5, 3)$ is the centre of that circle, then equation of the circle is (A) $(x-5)^2 + (y-3)^2 = 10^2$ (B) $(x-5)^2 + (y-3)^2 = 12^2$ (C) $(x-5)^2 + (y-3)^2 = 7^2$ (D) $(x-5)^2 + (y-3)^2 = 8^2$

   ›Reveal solutionSolution

   The line cuts the circle’s circumference in a 1:2 ratio, meaning it is a chord that subtends a central angle of $120^\circ$. The perpendicular distance from the centre to the chord, combined with the chord’s geometry, gives the radius. The radius is $10$, so the equation is $(x-5)^2+(y-3)^2=10^2$, option (A).

   The key idea: when a line divides the circumference of a circle in a given ratio, it is a chord that cuts the circle into two arcs whose lengths are in that ratio. The central angle subtended by the chord is proportional to the arc length. Here, the ratio $1:2$ means the smaller arc corresponds to a central angle of $120^\circ$ (since $1+2=3$ parts, and $360^\circ/3=120^\circ$ per part). The chord therefore subtends an angle of $120^\circ$ at the centre.

   Now, the perpendicular distance from the centre to the chord, the radius, and half the chord length form a right triangle. The angle at the centre between the radius and the perpendicular is half the subtended angle — here $60^\circ$. That lets us find the radius directly from the distance.

   1. Find the perpendicular distance from the centre to the line.

      The line is $4x+3y-4=0$. Centre $C(5,3)$.

      Distance $d = \frac{|4(5)+3(3)-4|}{\sqrt{4^2+3^2}} = \frac{|20+9-4|}{5} = \frac{25}{5} = 5$.

   2. Relate the distance to the radius.

      In the right triangle formed by the centre, the foot of the perpendicular, and one endpoint of the chord, the angle at the centre is $60^\circ$ (half of $120^\circ$).

      The perpendicular distance $d$ is the adjacent side to this $60^\circ$ angle, and the radius $R$ is the hypotenuse. …

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8. TG EAPCET 2022Set eng-2022-07-18-AN1 markMCQ
   Q.From a point $A(0, 3)$ on the circle $(x+2)^2 + (y-3)^2 = 4$, a chord $AB$ is drawn and it is extended to a point $Q$ such that $AQ = 2AB$. Then the locus of $Q$ is (A) $(x+4)^2 + (y-3)^2 = 16$ (B) $(x+1)^2 + (y-3)^2 = 32$ (C) $(x+1)^2 + (y-3)^2 = 4$ (D) $(x+1)^2 + (y-3)^2 = 1$

   ›Reveal solutionSolution

   $B$ is the midpoint of $AQ$, so $B=\tfrac{A+Q}{2}$. Substituting into the given circle and simplifying gives $(x+4)^2+(y-3)^2=16$ — option (A).

   The circle is $(x+2)^2+(y-3)^2=4$, centre $(-2,3)$, radius $2$. The point $A(0,3)$ lies on it since $(0+2)^2+(3-3)^2=4$.

   A chord $AB$ is extended to $Q$ with $AQ=2AB$. Since $Q$ lies on ray $AB$ at twice the distance, $B$ is the midpoint of $AQ$:

   $B=\frac{A+Q}{2}=\left(\frac{Q_x}{2},\ \frac{Q_y+3}{2}\right).$

   $B$ lies on the given circle, so $(B_x+2)^2+(B_y-3)^2=4$:

   $\left(\frac{Q_x}{2}+2\right)^2+\left(\frac{Q_y+3}{2}-3\right)^2=4.$ …

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9. TG EAPCET 2026Set eng-2026-05-11-FN1 markMCQ
   Q.A circle $S \equiv x^2 + y^2 + 4x + 2fy + c = 0$ passes through the centre of the circle $x^2 + y^2 - 4x + 6y - 2 = 0$. If the line $3x - 3y = c$ passes through the centre of the circle $S = 0$ then the length of the tangent drawn from the point $(1,1)$ to the circle $S = 0$ is (A) $8$ (B) $\sqrt{2}$ (C) $5$ (D) $\sqrt{10}$

   ›Reveal solutionSolution

   The key idea is to use the given conditions to find the unknown parameters $f$ and $c$ of circle $S$, then compute the length of the tangent from $(1,1)$ using the formula $\sqrt{S_1}$. The final length is $\sqrt{10}$.

   The problem gives us a circle $S$ with one unknown parameter $f$ and a constant $c$, and two conditions that let us determine both. Once we know the circle completely, the tangent length from a point is straightforward.

   Concept: A circle's centre is found by completing the square. The length of the tangent from an external point $(x_1, y_1)$ to a circle $x^2 + y^2 + 2gx + 2fy + c = 0$ is $\sqrt{x_1^2 + y_1^2 + 2gx_1 + 2fy_1 + c}$ — this is just the square root of the value obtained by substituting the point into the circle's equation.

   1. Find the centre of the given circle.

      The circle $x^2 + y^2 - 4x + 6y - 2 = 0$ can be rewritten by completing squares:

      $(x^2 - 4x) + (y^2 + 6y) = 2$

      $(x-2)^2 - 4 + (y+3)^2 - 9 = 2$

      $(x-2)^2 + (y+3)^2 = 15$

      So its centre is $C_1 = (2, -3)$.

   2. Circle $S$ passes through $C_1$.

      Circle $S$ is $x^2 + y^2 + 4x + 2fy + c = 0$. Its centre is $(-2, -f)$ (since $2g = 4 \Rightarrow g = 2$, and $2f$ is the coefficient of $y$, so centre is $(-g, -f) = (-2, -f)$).

      Since $C_1 = (2, -3)$ lies on $S$, substitute:

      $2^2 + (-3)^2 + 4(2) + 2f(-3) + c = 0$

      $4 + 9 + 8 - 6f + c = 0$

      $21 - 6f + c = 0$

      So $c = 6f - 21$. …(1)

   3. The line $3x - 3y = c$ passes through the centre of $S$.

      Centre of $S$ is $(-2, -f)$. Substitute into the line:

      $3(-2) - 3(-f) = c$

      $-6 + 3f = c$

      So $c = 3f - 6$. …(2)

   4. Solve for $f$ and $c$.

      From (1) and (2): $6f - 21 = 3f - 6$

      $3f = 15 \Rightarrow f = 5$

      Then $c = 3(5) - 6 = 9$. …

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10. TG EAPCET 2022Set eng-2022-07-19-AN1 markMCQ
    Q.If the circles $x^2 + y^2 - 16x - 20y + 164 = r^2 \ (r > 0)$ and $x^2 + y^2 - 8x - 14y + 29 = 0$ intersect in two distinct points, then the maximum possible integral value of $r$ is (A) $1$ (B) $10$ (C) $-2$ (D) $2$

    ›Reveal solutionSolution

    For two distinct intersection points, the distance between the centres must satisfy $|R_1 - R_2| < d < R_1 + R_2$. Here $R_1 = r$, $R_2 = 6$, and $d = 5$, which gives $1 < r < 11$. The maximum integer value satisfying this is $r = 10$, so option (B) is correct.

    ---

    The problem gives two circles and asks for the largest integer $r$ such that they intersect in two distinct points. The first circle has a parameter $r$ in its radius, and the second is fixed. The condition for two distinct intersection points is that the distance between the centers lies strictly between the difference and the sum of the radii.

    Let’s rewrite each circle in center-radius form.

    Circle 1: $x^2 + y^2 - 16x - 20y + 164 = r^2$

    Complete the square:

    • $x^2 - 16x = (x-8)^2 - 64$
    • $y^2 - 20y = (y-10)^2 - 100$

    So the equation becomes:

    $(x-8)^2 - 64 + (y-10)^2 - 100 + 164 = r^2$

    $(x-8)^2 + (y-10)^2 = r^2$

    Thus Circle 1 has center $C_1 = (8, 10)$ and radius $R_1 = r$ (given $r > 0$).

    Circle 2: $x^2 + y^2 - 8x - 14y + 29 = 0$

    Complete the square:

    • $x^2 - 8x = (x-4)^2 - 16$
    • $y^2 - 14y = (y-7)^2 - 49$

    So:

    $(x-4)^2 - 16 + (y-7)^2 - 49 + 29 = 0$

    $(x-4)^2 + (y-7)^2 = 36$

    Thus Circle 2 has center $C_2 = (4, 7)$ and radius $R_2 = 6$.

    Now compute the distance $d$ between the centers:

    $d = \sqrt{(8-4)^2 + (10-7)^2} = \sqrt{4^2 + 3^2} = \sqrt{16+9} = \sqrt{25} = 5$

    For two distinct intersection points, we need:

    $|R_1 - R_2| < d < R_1 + R_2$

    Substitute $R_1 = r$, $R_2 = 6$, $d = 5$:

    1. Right inequality: $d < R_1 + R_2$ → $5 < r + 6$ → $r > -1$. Since $r > 0$, this is automatically satisfied. …

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11. TG EAPCET 2025Set eng-2025-05-03-AN1 markMCQ
    Q.If $a = 3$, $b = 5$, $c = 7$ are the sides of a triangle ABC, then its circumradius is (A) $\dfrac{7}{\sqrt{3}}$ (B) $\dfrac{15}{2}$ (C) $\dfrac{15\sqrt{3}}{4}$ (D) $\dfrac{\sqrt{3}}{2}$

    ›Reveal solutionSolution

    The circumradius of a triangle with sides 3, 5, 7 is found using the formula $R = \frac{abc}{4\Delta}$, where $\Delta$ is the area via Heron’s formula. The result is $\frac{7}{\sqrt{3}}$, so the correct option is (A).

    The key idea: For any triangle, the circumradius $R$ relates the sides and the area. Instead of memorizing a separate formula, we combine the law of sines with the area formula. Here, the sides are given, so Heron’s formula gives the area, and then $R = \frac{abc}{4\Delta}$ does the rest.

    1. Find the semi-perimeter For sides $a=3$, $b=5$, $c=7$, the semi-perimeter is

    $s = \frac{a+b+c}{2} = \frac{3+5+7}{2} = \frac{15}{2}.$

    2. Compute the area using Heron’s formula Heron’s formula: $\Delta = \sqrt{s(s-a)(s-b)(s-c)}$. Substitute:

    $\Delta = \sqrt{\frac{15}{2} \left(\frac{15}{2}-3\right) \left(\frac{15}{2}-5\right) \left(\frac{15}{2}-7\right)}.$

    Simplify each factor:

    $s-a = \frac{15}{2} - 3 = \frac{9}{2}, \quad s-b = \frac{15}{2} - 5 = \frac{5}{2}, \quad s-c = \frac{15}{2} - 7 = \frac{1}{2}.$

    So

    $\Delta = \sqrt{\frac{15}{2} \cdot \frac{9}{2} \cdot \frac{5}{2} \cdot \frac{1}{2}} = \sqrt{\frac{15 \cdot 9 \cdot 5 \cdot 1}{16}} = \sqrt{\frac{675}{16}}.$

    Simplify: $675 = 25 \cdot 27 = 25 \cdot 9 \cdot 3$, so

    $\Delta = \frac{\sqrt{675}}{4} = \frac{15\sqrt{3}}{4}.$

    3. Apply the circumradius formula The standard formula: $R = \frac{abc}{4\Delta}$. …

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12. TG EAPCET 2023Set eng-2023-05-13-AN1 markMCQ
    Q.A circle $S = x^2 + y^2 + 2gx + 2fy + 4 = 0$ cuts the circle $x^2 + y^2 - 4x - 4y - 4 = 0$ orthogonally and makes an angle of $60^\circ$ with the circle $x^2 + y^2 + 4x + 4y + 4 = 0$. Then the radius of the circle $S = 0$ is (A) $4$ (B) $3$ (C) $5$ (D) $1$

    ›Reveal solutionSolution

    Orthogonality forces $f=-g$; imposing the $60^\circ$ intersection then gives radius $=4$.

    Let $S:\;x^2+y^2+2gx+2fy+4=0$, so its centre is $(-g,-f)$ and radius $r=\sqrt{g^2+f^2-4}$.

    Orthogonal to $x^2+y^2-4x-4y-4=0$: using $2g_1g_2+2f_1f_2=c_1+c_2$ with $(g_2,f_2,c_2)=(-2,-2,-4)$,

    $2g(-2)+2f(-2)=4+(-4)=0\;\Rightarrow\;g+f=0\;\Rightarrow\;f=-g.$

    Then $r^2=g^2+f^2-4=2g^2-4$, and the centre of $S$ is $(-g,\,g)$.

    Angle $60^\circ$ with $x^2+y^2+4x+4y+4=0$: this circle has centre $(-2,-2)$ and radius $r_2=\sqrt{4+4-4}=2$. The distance between centres:

    $d^2=(-g+2)^2+(g+2)^2=2g^2+8.$

    The angle of intersection $\theta$ satisfies …

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