Question

In the given figure, the equation of the larger circle is $x^2+y^2+4y-5=0$ 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. $x^2+y^2=2\sqrt{7}x+2y$
• D. None of these

Answer 1

Answer: (A)

We have $x^2+y^2+4y-5=0.$
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 $C_1{C}_2=4.$
$\Rightarrow \sqrt{h^2+(k+2{)}^2}=3+r_2=4\dots (1)$
$\Rightarrow r_2=1$
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 $(x-\sqrt{7}{)}^2+(y-1{)}^2=1$