If the circles x² + y² = 9 and x² + y² + 8y + c = 0 touch each other, then c is equal to
(a) 15
(b) − 15
(c) 16
(d) − 16

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Solution

(a) 15
The centre of the circle x² + y² = 9 is (0, 0).
Let us denote it by C₁.
The centre of the circle x² + y²+ 8y + c = 0 is (0, −4).
Let us denote it by C₂.

The radius of x² + y² = 9 is 3 units.

x² + y²+ 8y + 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.
∴ C₁C₂= PC₂ + PC₁

⇒ PC₂= 4 − 3 = 1
⇒ PC₂= 1 = $\sqrt{16 - c}$
⇒ c = 15

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