Question

Let \[ S\equiv x^2+y^2-6x+4y+c=0,\qquad S'\equiv x^2+y^2-4x+6y+9=0, \] \[ S''\equiv x^2+y^2+5x+3y+k=0 \] be three circles. If the angles of intersection of the circle \(S'=0\) with the circles \(S=0\) and \(S''=0\) are respectively \[ \frac{\pi}{4} \quad\text{and}\quad \frac{\pi}{2}, \] then \(c+k=\)

• A. \(1\)
• B. \(15\)
• C. \(21\)
• D. \(9\)

Hint:

For two intersecting circles, \[ \boxed{ \cos\theta = \frac{r_1^2+r_2^2-d^2}{2r_1r_2} } \] where \(d\) is the distance between their centres. For orthogonal circles, \[ \boxed{r_1^2+r_2^2=d^2.} \]

Answer 1

The Correct Option is A

Step 1: Find the radii of the circles. For \[ S:x^2+y^2-6x+4y+c=0, \] the centre is \[ (3,-2), \] and \[ r_1^2=13-c. \] For \[ S':x^2+y^2-4x+6y+9=0, \] the centre is \[ (2,-3), \] and \[ r_2^2=4. \] For \[ S'':x^2+y^2+5x+3y+k=0, \] the centre is \[ \left(-\frac52,-\frac32\right), \] and \[ r_3^2=\frac{17}{2}-k. \]

Step 2: Use the angle of intersection formula. The distance between the centres of \[ S \] and \[ S' \] is \[ d^2=(3-2)^2+(-2+3)^2=2. \] Using \[ \cos\theta = \frac{r_1^2+r_2^2-d^2}{2r_1r_2}, \] with \[ \theta=\frac{\pi}{4}, \] we get \[ \frac{(13-c)+4-2}{4\sqrt{13-c}} = \frac1{\sqrt2}. \] Solving, \[ 13-c=8, \] so \[ c=5. \] For \[ S' \] and \[ S'', \] the distance between centres is \[ d^2=\left(2+\frac52\right)^2+\left(-3+\frac32\right)^2=\frac{45}{2}. \] Since the angle of intersection is \[ \frac{\pi}{2}, \] \[ r_2^2+r_3^2=d^2. \] Hence, \[ 4+\frac{17}{2}-k = \frac{45}{2}, \] which gives \[ k=-4. \]

Step 3: Find the required sum. Therefore, \[ c+k = 5+(-4) = 1. \] Hence, \[ \boxed{1}. \] Thus, \[ \boxed{(A)} \] is the correct answer.