If the circles $x^{2} + y^{2} = 9$ and $x^{2} + y^{2} + + 8 y + c = 0$
touch each other, then c is equal to

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Solution

The correct option is A
15

The centre of the circle $x^{2} + y^{2} = 9$ is (0, 0).

Let us denote it by $C_{1}$ .

The centre of the circle $x^{2} + y^{2} + 8 y + c = 0$ is (0, -4).

Let us denote it by $C_{2} .$

The radius $x^{2} + y^{2} = 9$ is 3 units.

$x^{2} + y^{2} + 8 y + c = 0$

$\Rightarrow (x - 0)^{2} + (y + 4)^{2} = 16 - c = (\sqrt{16 - c})^{2}$

Therefore, the radius of the above circle is
$\sqrt{16 - c}$

Let the circles touch each other at P.

$∴ C_{1} C_{2} = P C_{2} + P C_{1}$

$\Rightarrow P C_{2} = 4 - 3 = 1$

$\Rightarrow P C_{2} = 1 = \sqrt{16 - C}$

$\Rightarrow c = 15$

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