Question

If the circle $x^{2} + y^{2} = 4$ bisects the circumference of the circle$x^{2} + y^{2} - 2x + 6y + a = 0,$ then a equals

• A. 4
• B. – 4
• C. 16
• D. – 16

Answer 1

Answer: (C)

16

Answer 2

The common chord of given circles is $S_{1} - S_{2} = 0$

$\Rightarrow 2x - 6y - 4 - a = 0$ …..(i)

Since, $x^{2} + y^{2} = 4$ bisects the circumferences of the circle $x^{2} + y^{2} - 2x + 6y + a = 0,$ therefore (i) passes through the centre of second circle i.e. (1, – 3).

$\therefore$ 2 + 18 – 4 – a = 0

⇒ a = 16.