Question

In the given figure, the equation of the larger circle is $x^{2}+y^{2}+4 y-5=0$ and the distance between centres is $4$ . Then the equation of smaller circle is

• A. $(x-\sqrt{7})^{2}+(y-1)^{2}=1$
• B. $(x+\sqrt{7})^{2}+(y-1)^{2}=1$
• C. $x^{2}+y^{2}=2 \sqrt{7} x+2 y$
• D. None of these

Answer 1

Answer: (A)

We have $x^{2}+y^{2}+4 y-5=0 .$
Its centre is $C_{1}(0,-2), r_{1}=\sqrt{4+5}=3 .$
Let $C_{2}(h, k)$ be the centre of the smaller circle and its radius $r_{2} .$
Then $C_{1} C_{2}=4 .$
$\Rightarrow \sqrt{h^{2}+(k+2)^{2}}=3+r_{2}=4 \ldots(1)$
$\Rightarrow r_{2}=1$
But $k=r_{2}=1$ [it touches $x$-axis
$\therefore $ From eq $(1), 4=\sqrt{h^{2}+(1+2)^{2}}$
$\Rightarrow 16=h^{2}+9 \Rightarrow h^{2}=7 \Rightarrow h=\pm \sqrt{7}$
Since $h>0$
$\therefore h=\sqrt{7}$
Hence, required circle is $(x-\sqrt{7})^{2}+(y-1)^{2}=1$