Circle JEE Mathematics Worksheets Set 01

Question. A circle \( S = 0 \) is drawn with its centre at \( (-1, 1) \) so as to touch the circle \( x^2 + y^2 - 4x + 6y - 3 = 0 \) externally. Find the intercept made by the circle \( S = 0 \) on the coordinate axes.
Answer: \( S = 0 \), centre \( C_1 \equiv (-1, 1) \), radius \( r_1 \).
\( S_1 \equiv x^2 + y^2 - 4x + 6y - 3 = 0 \).
Centre \( C_2 \equiv (2, -3) \), radius \( r_2 = 4 \).
\( S_1 \) & \( S_2 \) touches externally
\( \therefore C_1 C_2 = r_1 + r_2 \)
\( \implies \sqrt{(2 + 1)^2 + (-3 - 1)^2} = r_1 + 4 \)
\( \implies r_1 = 5 - 4 = 1 \)
\( \therefore \) Circle is \( S \equiv (x + 1)^2 + (y - 1)^2 = 1^2 \)
\( \implies S \equiv x^2 + y^2 + 2x - 2y + 1 = 0 \)
\( \therefore \) x-intercept \( = 2\sqrt{g^2 - c} = 2\sqrt{1 - 1} = 0 \)
y-intercept \( = 2\sqrt{f^2 - c} = 2\sqrt{1 - 1} = 0 \)

Question. Find the equation to the circle which is such that the length of the tangents to it from the points (1, 0), (2, 0) and (3, 2) are 1, \( \sqrt{7} \), \( \sqrt{2} \) respectively.
Answer: Let circle \( S \equiv x^2 + y^2 + 2gx + 2fy + c = 0 \).
Given \( \sqrt{S_1} = 1, \sqrt{S_2} = \sqrt{7}, \sqrt{S_3} = \sqrt{2} \)
\( \implies \sqrt{1^2 + 0^2 + 2g(1) + 2f(0) + c} = 1 \)
\( \implies 2g + c + 1 = 1 \implies 2g = -c \dots(1) \)
& \( \sqrt{2^2 + 0^2 + 2g(2) + 2f(0) + c} = \sqrt{7} \)
\( \implies 4 + 4g + c = 7 \implies 4g + c = 3 \dots(2) \)
& \( \sqrt{3^2 + 2^2 + 2g(3) + 2f(2) + c} = \sqrt{2} \)
\( \implies 6g + 4f + c + 13 = 2 \implies 6g + 4f + c = -11 \dots(3) \)
By (1) & (2) \( \implies c = -3 \) & \( g = \frac{3}{2} \)
Put c & g in (3) we get \( \implies f = -\frac{17}{4} \)
So equation of circle is \( x^2 + y^2 + 3x - \frac{17}{2}y - 3 = 0 \)
\( \implies 2(x^2 + y^2) + 6x - 17y - 6 = 0 \)

Question. The coordinates of the point P if the triangle AQB has the maximum area.
Answer: Area \( \Delta AQB = \frac{1}{2} \times 10 \times 5 \sin\theta = 25 \sin\theta \).
Area max. at \( \theta = 90^\circ \implies A, Q, P \) are collinear.
\( \therefore m_{AQ} = 1 \implies \frac{k-5}{11} = 1 \implies k = 16 \).
\( \implies y = 16 \) so coordinate of P (11, 16).

Question. The coordinates of the point P if Q is the middle point of AP.
Answer: \( \frac{11 - 5}{2} = 5 \cos\theta \implies \cos\theta = \frac{3}{5} \)
\( \frac{k + 0}{2} = 5 \sin\theta \implies k = 10 \sin\theta = 10 \cdot \frac{4}{5} = 8 \).
\( \sin\theta = \frac{4}{5} \) (\( \because \) P lies above x-axis). \( \therefore P(11, 8) \).

Question. The coordinates of P if the area of the triangle AQB is \( (1/4)^{\text{th}} \) of the area of the triangle APC.
Answer: \( \Delta AQB = \frac{1}{4} \Delta APC \)
\( \implies 25 \sin\theta = \frac{1}{4} \cdot \frac{1}{2} \times 16 \times k \implies k = \frac{25}{2} \sin\theta \).
\( m_{AQ} = m_{AP} \implies \frac{5\sin\theta}{5(\cos\theta+1)} = \frac{k}{16} \)
\( \implies \frac{\sin\theta}{\cos\theta+1} = \frac{25}{2.16} \sin\theta \implies \cos\theta = \frac{32}{25} - 1 = \frac{7}{25} \).
\( \sin\theta = \frac{24}{25} \therefore k = \frac{25}{2} \times \frac{24}{25} \implies k = 12 \therefore P(11, 12) \).

Question. Find the locus of the mid point of the chord of a circle \( x^2 + y^2 = 4 \) such that the segment intercepted by the chord on the curve \( x^2 - 2x - 2y = 0 \) subtends a right angle at the origin.
Answer: Let mid point be \( P(h, k) \). Chord \( T = S_1 \implies hx + ky = h^2 + k^2 \).
Homogenization: \( x^2 - 2(x+y)\left(\frac{hx+ky}{h^2+k^2}\right) = 0 \).
For right angle, coeff of \( x^2 + \) coeff of \( y^2 = 0 \).
\( \implies 1 - \frac{2h}{h^2+k^2} - \frac{2k}{h^2+k^2} = 0 \implies h^2+k^2-2h-2k=0 \).
Locus is \( x^2+y^2-2x-2y=0 \).

Question. Let K denotes the square of the diameter of the circle whose diameter is the common chord of the two circles \( x^2 + y^2 + 2x + 3y + 1 = 0 \) and \( x^2 + y^2 + 4x + 3y + 2 = 0 \) and W denotes the sum of the abscissa and ordinate of a point P where all variable chords of the curve \( y^2 = 8x \) subtending right angles at the origin, are concurrent. and H denotes the square of the length of the tangent from the point (3, 0) on the circle \( 2x^2 + 2y^2 + 5y - 16 = 0 \). Find the value of KWH.
Answer: Radical axis: \( 2x + 1 = 0 \implies x = -\frac{1}{2} \).
\( C_1M = 1/2 \), \( AM = \sqrt{2} \implies AB = 2\sqrt{2} \implies K = (2\sqrt{2})^2 = 8 \).
For \( y^2=8x \), concurrent point is (8, 0) \( \implies W = 8+0=8 \).
For circle \( x^2+y^2+\frac{5}{2}y-8=0 \), at (3, 0), \( H = 9+0-8=1 \).
\( KWH = 8 \times 8 \times 1 = 64 \).

Question. Find the locus of the centre of a circle which cuts two given circles orthogonally.
Answer: Let circles be \( S_1 = 0 \) and \( S_2 = 0 \).
Orthogonal condition: \( 2gg_1 + 2ff_1 = c + c_1 \) and \( 2gg_2 + 2ff_2 = c + c_2 \).
Subtracting: \( 2g(g_1 - g_2) + 2f(f_1 - f_2) = c_1 - c_2 \).
Locus of centre \( (-g, -f) \) is \( 2x(g_1 - g_2) + 2y(f_1 - f_2) + c_1 - c_2 = 0 \), which is the radical axis.
Example for \( x^2 + y^2 + 4x - 6y + 9 = 0 \) and \( x^2 + y^2 - 5x + 4y + 2 = 0 \):
Locus is \( 9x - 10y + 7 = 0 \).

Question. The straight line \( 2x - 3y = 1 \) divides the circular region \( x^2 + y^2 \le 6 \) into two parts. If \( S = \left\{ \left(2, \frac{3}{4}\right), \left(\frac{5}{2}, \frac{3}{4}\right), \left(\frac{1}{4}, -\frac{1}{4}\right), \left(\frac{1}{8}, \frac{1}{4}\right) \right\} \), then the number of point(s) in S lying inside the smaller part is
Answer: Origin lies left to the line. Points (2, 3/4) & (1/4, -1/4) lie in the smaller part & also in the circle so only two points.

Question. Two parallel chords of a circle of radius 2 are at a distance \( \sqrt{3} + 1 \) apart. If the chords subtend at the center, angles of \( \frac{\pi}{k} \) and \( \frac{2\pi}{k} \), where k > 0, then the value of [k] is
{Note : [k] denotes the largest integer less than or equal to k}
Answer: \( OM = 2\cos\frac{\pi}{2k} \) & \( ON = 2\cos\frac{\pi}{k} \)
\( OM + ON = \sqrt{3} + 1 \)
\( 2\left(\cos\frac{\pi}{2k} + \cos\frac{\pi}{k}\right) = \sqrt{3} + 1 \)
Let \( \frac{\pi}{k} = \theta \)
\( 2\left(\cos\frac{\theta}{2} + \cos\theta\right) = \sqrt{3} + 1 \)
\( \implies 2\cos\frac{\theta}{2} + 4\cos^2\frac{\theta}{2} - 2 = \sqrt{3} + 1 \)
\( \implies 4\cos^2\frac{\theta}{2} + 2\cos\frac{\theta}{2} - (\sqrt{3} + 3) = 0 \)
\( \cos\frac{\theta}{2} = \frac{-2 \pm \sqrt{4 + 16(\sqrt{3} + 3)}}{4} \)
\( = \frac{-1 \pm \sqrt{1 + 2 \cdot 1 \cdot 2\sqrt{3} + (2\sqrt{3})^2}}{2} \)
\( = \frac{-1 \pm (2\sqrt{3} + 1)^2}{2} = -\frac{2(\sqrt{3} + 1)}{2} \) or \( \frac{\sqrt{3}}{2} \)
\( = -(\sqrt{3} + 1) \) or \( \frac{\sqrt{3}}{2} \)
\( \cos\frac{\theta}{2} \neq -(\sqrt{3} + 1) \) since magnitude > 1
\( \therefore \cos\frac{\theta}{2} = \frac{\sqrt{3}}{2} \)
\( \implies \frac{\theta}{2} = \frac{\pi}{6} \)
\( \implies \theta = \frac{\pi}{3} = \frac{\pi}{k} \)
\( \implies k = 3 \)

Question. Line \( 2x + 3y + 1 = 0 \) is a tangent to a circle at \( (1, -1) \). This circle is orthogonal to a circle which is drawn having diameter as a line segment with end points \( (0, -1) \) and \( (-2, 3) \). Find equation of circle.
Answer: \( x(x + 2) + (y + 1) (y - 3) = 0 \)
\( x^2 + y^2 + 2x - 2y - 3 = 0 \)
& \( x^2 + y^2 + 2gx + 2fy + c = 0 \) are orthogonal
\( 2g + 2f (-1) = c - 3 \)
\( 2g - 2f + 3 = c \) ....(i)
circle passing through (1, -1)
\( (2g - 2f) + c + 2 = 0 \)
\( \implies c - 3 + c + 2 = 0 \)
\( \implies c = \frac{1}{2} \)
tangent at (1, -1)
\( x - y + 2(x + 1) + f(y - 1) + c = 0 \)
\( \implies x(g + 1) + y(f - 1) + g - f + c = 0 \)
\( \implies x(g + 1) + y(f - 1) + \frac{c}{2} + \frac{3}{2} + c = 0 \)
\( \implies x(g + 1) + y(f - 1) - \frac{3}{4} + c = 0 \)
& \( 2x + 3y + 1 = 0 \) are same tangent
\( \frac{g + 1}{2} = \frac{f - 1}{3} = \frac{-3/4}{1} \)
\( \implies g = \frac{-5}{2} \) & \( f = -\frac{5}{4} \)
circle is
\( x^2 + y^2 - 5x - \frac{5}{2}y + \frac{1}{2} = 0 \)
\( \implies 2x^2 + 2y^2 - 10x - 5y + 1 = 0 \)

Question. (b) Let \( 2x^2 + y^2 - 3xy = 0 \) be the equation of a pair of tangents drawn from the origin 'O' to a circle of radius 3 with centre in the first quadrant. If A is one of the points of contact, find the length of OA.
Answer: Pair of lines \( 2x^2 - 3xy + y^2 = 0 \)
\( \tan\theta = \frac{2\sqrt{\frac{9}{4} - 2}}{3} = \frac{1}{3} \)
\( \sin\theta = \frac{1}{\sqrt{10}} \), \( \cos\theta = \frac{3}{\sqrt{10}} \)
\( \tan\left(\frac{\theta}{2}\right) = \frac{1 - \cos\theta}{\sin\theta} = \frac{1 - \frac{3}{\sqrt{10}}}{\frac{1}{\sqrt{10}}} = \sqrt{10} - 3 \)
In \( \Delta OAC \)
\( \tan\frac{\theta}{2} = \frac{3}{OA} \)
\( \implies OA = \frac{3}{\tan\left(\frac{\theta}{2}\right)} = \frac{3}{\sqrt{10} - 3} \times \frac{\sqrt{10} + 3}{\sqrt{10} + 3} = 3(\sqrt{10} + 3) \).

Question. (a) Find the equation of the circle which passes through the points of intersection of circles \( x^2 + y^2 - 2x - 6y + 6 = 0 \) and \( x^2 + y^2 + 2x - 6y + 6 = 0 \) and intersects the circle \( x^2 + y^2 + 4x + 6y + 4 = 0 \) orthogonally.
Answer: \( (x^2 + y^2 - 2x - 6y + 6) + \lambda (x^2 + y^2 + 2x - 6y + 6) = 0 \)
\( (1 + \lambda) (x^2 + y^2) + 2x (\lambda - 1) - 6y (\lambda + 1) + 6 (\lambda + 1) = 0 \)
\( \implies x^2 + y^2 + 2x\frac{(\lambda - 1)}{(\lambda + 1)} - 6y + 6 = 0 \)
Cuts orthogonally \( x^2 + y^2 + 4x + 6y + 4 = 0 \)
\( 2 \cdot \frac{(\lambda - 1)}{(\lambda + 1)} \cdot 2 + 2 \cdot (-3) (3) = 6 + 4 \)
\( \implies \frac{4(\lambda - 1)}{(\lambda + 1)} - 18 = 10 \)
\( \implies \frac{2(\lambda - 1)}{(\lambda + 1)} - 9 = 5 \)
\( \implies \frac{\lambda - 1}{\lambda + 1} = 7 \)
Equation: \( x^2 + y^2 + 14x - 6y + 6 = 0 \).

Question. (b) Tangents TP and TQ are drawn from a point T to the circle \( x^2 + y^2 = a^2 \). If the point T lies on the line \( px + qy = r \), find the locus of centre of the circumcircle of triangle TPQ.
Answer: C is mid point of OT
\( 2(CT) = OT \)
\( \frac{2|Ph + qk - r|}{\sqrt{p^2 + q^2}} = \frac{|-r|}{\sqrt{p^2 + q^2}} \)
C & O lie same side of line
\( \therefore 2(Ph + qk - r) = -r \)
Locus is \( 2px + 2qy = r \).

Question. (a) Extremities of a diagonal of a rectangle are (0, 0) & (4, 3). Find the equation of the tangents to the circumcircle of a rectangle which are parallel to this diagonal.
Answer: Centre \( \left(2, \frac{3}{2}\right) \), \( r = \frac{5}{2} \)
\( m_{AB} = \frac{3}{4} \)
\( \therefore \) Tangent \( 3x - 4y + \lambda = 0 \)
\( \frac{|6 - 6 + \lambda|}{5/2} = \frac{5}{2} \)
\( \implies \frac{| \lambda |}{5} = \frac{5}{2} \)
\( \implies | \lambda | = \frac{25}{2} \)
\( \implies \lambda = \pm \frac{25}{2} \)
Tangents are \( 6x - 8y \pm 25 = 0 \).

Question. (b) Find the point on the straight line, \( y = 2x + 11 \) which is nearest to the circle, \( 16(x^2 + y^2) + 32x - 8y - 50 = 0 \).
Answer: \( 16(x^2 + y^2) + 32x - 8y - 50 = 0 \)
\( \implies x^2 + y^2 + 2x - \frac{1}{2}y - \frac{25}{8} = 0 \)
centre \( \left(-1, \frac{1}{4}\right) \)
Line \( y = 2x + 11 \)
\( \implies 2x - y + 11 = 0 \)
\( \perp \) line is \( x + 2y + \lambda = 0 \)
passing through centre \( \left(-1, \frac{1}{4}\right) \)
\( \implies -1 + \frac{1}{2} + \lambda = 0 \)
\( \implies \lambda = \frac{1}{2} \)
\( \perp \) line is \( 2x + 4y + 1 = 0 \)
Intersection point M is \( \left(-\frac{9}{2}, 2\right) \).

Question. (c) A circle of radius 2 units rolls on the outerside of the circle, \( x^2 + y^2 + 4x = 0 \), touching it externally. Find the locus of the centre of this outer circle. Also find the equations of the common tangents of the two circles when the line joining the centres of the two circles is inclined at an angle of \( 60^\circ \) with x-axis.
Answer: \( S \equiv x^2 + y^2 + 4x = 0 \)
centre (-2, 0), r = 2
circle is
\( (x + 2)^2 + y^2 = 4^2 \)
\( x^2 + y^2 + 4x - 12 = 0 \)
Line passing (-2, 0) & \( \theta = 60^\circ \)
\( \frac{x + 2}{1/2} = \frac{y}{\sqrt{3}/2} = \pm 4 \)
\( x = \pm 2 - 2, y = \pm 2\sqrt{3} \)
\( P(0, 2\sqrt{3}) \) & \( Q(-4, -2\sqrt{3}) \)
D.C.T. tangent parallel to line joining centres.
(\( \because \) radius is equal to both circles)
\( \sqrt{3}x - y + \lambda = 0 \)
\( 2 = \frac{| -2\sqrt{3} + \lambda |}{2} \)
\( \implies \lambda = 2\sqrt{3} \pm 4 \)
\( \sqrt{3}x - y + 2\sqrt{3} \pm 4 = 0 \)
T.C.T. \( \perp \) lines passes M & N
M & N are \( (-1, \sqrt{3}) \) & \( (-3, -\sqrt{3}) \)
\( x + \sqrt{3}y + \mu = 0 \)
\( -1 + 3 + \mu = 0 \)
\( \implies \mu = -2 \)
\( \implies x + \sqrt{3}y - 2 = 0 \)
& \( -3 - 3 + \mu = 0 \)
\( \implies \mu = 6 \)
\( \implies x + \sqrt{3}y + 6 = 0 \)

Question. (a) Extremities of a diagonal of a rectangle are (0, 0) & (4, 3). Find the equation of the tangents to the circumcircle of a rectangle which are parallel to this diagonal.
Answer: Centre \( \left(2, \frac{3}{2}\right) \), \( r = \frac{5}{2} \)
\( m_{AB} = \frac{3}{4} \)
\( \therefore \) Tangent \( 3x - 4y + \lambda = 0 \)
\( \frac{|6 - 6 + \lambda|}{5/2} = \frac{5}{2} \)
\( \implies \frac{| \lambda |}{5} = \frac{5}{2} \)
\( \implies | \lambda | = \frac{25}{2} \)
\( \implies \lambda = \pm \frac{25}{2} \)
Tangents are \( 6x - 8y \pm 25 = 0 \).

Question. (b) Find the point on the straight line, \( y = 2x + 11 \) which is nearest to the circle, \( 16(x^2 + y^2) + 32x - 8y - 50 = 0 \).
Answer: \( 16(x^2 + y^2) + 32x - 8y - 50 = 0 \)
\( \implies x^2 + y^2 + 2x - \frac{1}{2}y - \frac{25}{8} = 0 \)
centre \( \left(-1, \frac{1}{4}\right) \)
Line \( y = 2x + 11 \)
\( \implies 2x - y + 11 = 0 \)
\( \perp \) line is \( x + 2y + \lambda = 0 \)
passing through centre \( \left(-1, \frac{1}{4}\right) \)
\( \implies -1 + \frac{1}{2} + \lambda = 0 \)
\( \implies \lambda = \frac{1}{2} \)
\( \perp \) line is \( 2x + 4y + 1 = 0 \)
Intersection point M is \( \left(-\frac{9}{2}, 2\right) \).

Question. (c) A circle of radius 2 units rolls on the outerside of the circle, \( x^2 + y^2 + 4x = 0 \), touching it externally. Find the locus of the centre of this outer circle. Also find the equations of the common tangents of the two circles when the line joining the centres of the two circles is inclined at an angle of \( 60^\circ \) with x-axis.
Answer: \( S \equiv x^2 + y^2 + 4x = 0 \)
centre (-2, 0), r = 2
circle is
\( (x + 2)^2 + y^2 = 4^2 \)
\( x^2 + y^2 + 4x - 12 = 0 \)
Line passing (-2, 0) & \( \theta = 60^\circ \)
\( \frac{x + 2}{1/2} = \frac{y}{\sqrt{3}/2} = \pm 4 \)
\( x = \pm 2 - 2, y = \pm 2\sqrt{3} \)
\( P(0, 2\sqrt{3}) \) & \( Q(-4, -2\sqrt{3}) \)
D.C.T. tangent parallel to line joining centres.
(\( \because \) radius is equal to both circles)
\( \sqrt{3}x - y + \lambda = 0 \)
\( 2 = \frac{| -2\sqrt{3} + \lambda |}{2} \)
\( \implies \lambda = 2\sqrt{3} \pm 4 \)
\( \sqrt{3}x - y + 2\sqrt{3} \pm 4 = 0 \)
T.C.T. \( \perp \) lines passes M & N
M & N are \( (-1, \sqrt{3}) \) & \( (-3, -\sqrt{3}) \)
\( x + \sqrt{3}y + \mu = 0 \)
\( -1 + 3 + \mu = 0 \)
\( \implies \mu = -2 \)
\( \implies x + \sqrt{3}y - 2 = 0 \)
& \( -3 - 3 + \mu = 0 \)
\( \implies \mu = 6 \)
\( \implies x + \sqrt{3}y + 6 = 0 \)

Question. (b) Let \( 2x^2 + y^2 - 3xy = 0 \) be the equation of a pair of tangents drawn from the origin 'O' to a circle of radius 3 with centre in the first quadrant. If A is one of the points of contact, find the length of OA.
Answer: Pair of lines \( 2x^2 - 3xy + y^2 = 0 \)
\( \tan\theta = \frac{2\sqrt{\frac{9}{4} - 2}}{3} = \frac{1}{3} \)
\( \sin\theta = \frac{1}{\sqrt{10}} \), \( \cos\theta = \frac{3}{\sqrt{10}} \)
\( \tan\left(\frac{\theta}{2}\right) = \frac{1 - \cos\theta}{\sin\theta} = \frac{1 - \frac{3}{\sqrt{10}}}{\frac{1}{\sqrt{10}}} = \sqrt{10} - 3 \)
In \( \Delta OAC \)
\( \tan\frac{\theta}{2} = \frac{3}{OA} \)
\( \implies OA = \frac{3}{\tan\left(\frac{\theta}{2}\right)} = \frac{3}{\sqrt{10} - 3} \times \frac{\sqrt{10} + 3}{\sqrt{10} + 3} = 3(\sqrt{10} + 3) \).

Question. (a) Find the equation of the circle which passes through the points of intersection of circles \( x^2 + y^2 - 2x - 6y + 6 = 0 \) and \( x^2 + y^2 + 2x - 6y + 6 = 0 \) and intersects the circle \( x^2 + y^2 + 4x + 6y + 4 = 0 \) orthogonally.
Answer: \( (x^2 + y^2 - 2x - 6y + 6) + \lambda (x^2 + y^2 + 2x - 6y + 6) = 0 \)
\( (1 + \lambda) (x^2 + y^2) + 2x (\lambda - 1) - 6y (\lambda + 1) + 6 (\lambda + 1) = 0 \)
\( \implies x^2 + y^2 + 2x\frac{(\lambda - 1)}{(\lambda + 1)} - 6y + 6 = 0 \)
Cuts orthogonally \( x^2 + y^2 + 4x + 6y + 4 = 0 \)
\( 2 \cdot \frac{(\lambda - 1)}{(\lambda + 1)} \cdot 2 + 2 \cdot (-3) (3) = 6 + 4 \)
\( \implies \frac{4(\lambda - 1)}{(\lambda + 1)} - 18 = 10 \)
\( \implies \frac{2(\lambda - 1)}{(\lambda + 1)} - 9 = 5 \)
\( \implies \frac{\lambda - 1}{\lambda + 1} = 7 \)
Equation: \( x^2 + y^2 + 14x - 6y + 6 = 0 \).

Question. Line \( 2x + 3y + 1 = 0 \) is a tangent to a circle at \( (1, -1) \). This circle is orthogonal to a circle which is drawn having diameter as a line segment with end points \( (0, -1) \) and \( (-2, 3) \). Find equation of circle.
Answer: \( x(x + 2) + (y + 1) (y - 3) = 0 \)
\( x^2 + y^2 + 2x - 2y - 3 = 0 \)
& \( x^2 + y^2 + 2gx + 2fy + c = 0 \) are orthogonal
\( 2g + 2f (-1) = c - 3 \)
\( 2g - 2f + 3 = c \) ....(i)
circle passing through (1, -1)
\( (2g - 2f) + c + 2 = 0 \)
\( \implies c - 3 + c + 2 = 0 \)
\( \implies c = \frac{1}{2} \)
tangent at (1, -1)
\( x - y + 2(x + 1) + f(y - 1) + c = 0 \)
\( \implies x(g + 1) + y(f - 1) + g - f + c = 0 \)
\( \implies x(g + 1) + y(f - 1) + \frac{c}{2} + \frac{3}{2} + c = 0 \)
\( \implies x(g + 1) + y(f - 1) - \frac{3}{4} + c = 0 \)
& \( 2x + 3y + 1 = 0 \) are same tangent
\( \frac{g + 1}{2} = \frac{f - 1}{3} = \frac{-3/4}{1} \)
\( \implies g = \frac{-5}{2} \) & \( f = -\frac{5}{4} \)
circle is
\( x^2 + y^2 - 5x - \frac{5}{2}y + \frac{1}{2} = 0 \)
\( \implies 2x^2 + 2y^2 - 10x - 5y + 1 = 0 \)

Question. (b) The centres of two circles \( C_1 \) and \( C_2 \) each of unit radius are at a distance of 6 units from each other. Let P be the mid point of the line segment joining the centres of \( C_1 \) and \( C_2 \) and C be a circle touching circles \( C_1 \) and \( C_2 \) externally. If a common tangent to \( C_1 \) and C passing through P is also a common tangent to \( C_2 \) and C, then the radius of the circle C is
Answer: In \( \Delta PMQ \)
\( PQ^2 = (PM)^2 + r^2 \)
\( = 8 + r^2 \)
\( r^2 + 2r + 1 - 9 = 8 + r^2 \)
\( 2r = 16 \)
\( r = 8 \)

Question. Two parallel chords of a circle of radius 2 are at a distance \( \sqrt{3} + 1 \) apart. If the chords subtend at the center, angles of \( \frac{\pi}{k} \) and \( \frac{2\pi}{k} \), where k > 0, then the value of [k] is
{Note : [k] denotes the largest integer less than or equal to k}
Answer: \( OM = 2\cos\frac{\pi}{2k} \) & \( ON = 2\cos\frac{\pi}{k} \)
\( OM + ON = \sqrt{3} + 1 \)
\( 2\left(\cos\frac{\pi}{2k} + \cos\frac{\pi}{k}\right) = \sqrt{3} + 1 \)
Let \( \frac{\pi}{k} = \theta \)
\( 2\left(\cos\frac{\theta}{2} + \cos\theta\right) = \sqrt{3} + 1 \)
\( \implies 2\cos\frac{\theta}{2} + 4\cos^2\frac{\theta}{2} - 2 = \sqrt{3} + 1 \)
\( \implies 4\cos^2\frac{\theta}{2} + 2\cos\frac{\theta}{2} - (\sqrt{3} + 3) = 0 \)
\( \cos\frac{\theta}{2} = \frac{-2 \pm \sqrt{4 + 16(\sqrt{3} + 3)}}{4} \)
\( = \frac{-1 \pm \sqrt{1 + 2 \cdot 1 \cdot 2\sqrt{3} + (2\sqrt{3})^2}}{2} \)
\( = \frac{-1 \pm (2\sqrt{3} + 1)^2}{2} = -\frac{2(\sqrt{3} + 1)}{2} \) or \( \frac{\sqrt{3}}{2} \)
\( = -(\sqrt{3} + 1) \) or \( \frac{\sqrt{3}}{2} \)
\( \cos\frac{\theta}{2} \neq -(\sqrt{3} + 1) \) since magnitude > 1
\( \therefore \cos\frac{\theta}{2} = \frac{\sqrt{3}}{2} \)
\( \implies \frac{\theta}{2} = \frac{\pi}{6} \)
\( \implies \theta = \frac{\pi}{3} = \frac{\pi}{k} \)
\( \implies k = 3 \)