Question

The centre of a circle passing through the points $(0, 0),(1, 0)$ and touching the circle $x^2 + y^2 = 9$ is

• A. $(3/2, \, 1/2)$
• B. $(1/2, \, 3/2)$
• C. $(1/2, \, 1/2)$
• D. $(1/2, \, 12^{1/2})$

Answer 1

Answer: (D)

$(1/2, \, 12^{1/2})$

Answer 2

Let $C_1 (h , k)$ be the centre of the required circle. Then,
$\sqrt{(h-0)^2 + (k-0)^2} = \sqrt{(h-1)^2 + (k-0)^2}$
$\, \, \, \, \sqrt{(h-0)^2 + (k-0)^2} = \sqrt{(h-1)^2 + (k-0)^2}$
$\Rightarrow \, \, \, \, \, \, h^2 + k^2 = h^2 - 2h + 1 + k^2$
$\Rightarrow \, \, \, \, \, \, -2h + 1 = 0 \Rightarrow h = 1/2$
Since, (0, 0) and (1, 0) lie inside the circle $x^2+ y^2 =9$.
Therefore, the required circle can touch the given circle
internally.
$i.e. \, \, \, \, \, \, \, \, \, C_1. C_2 = r_1 \sim r_2$
$\Rightarrow \, \, \, \, \, \, \sqrt{h^2 + k^2} = 3 - \sqrt{h^2 + K^2}$
$\Rightarrow \, \, \, \, \, \, 2\sqrt{h^2 + k^2} = 3 \Rightarrow 2\sqrt{\frac{1}{4} + K^2 = 3}$
$\Rightarrow \, \, \, \, \, \, \sqrt{\frac{1}{4} + k^2} = \frac{3}{2} \Rightarrow \frac{1}{4} + k^2 = \frac{9}{4}$
$\Rightarrow \, \, \, \, \, \, \, k^2 = 2 \Rightarrow k = \pm \sqrt2$