Question

If the circle $x^2+y^2=9$ touches the circle $x^2+y^2+6y+c=0$, then $c$ is equal to-

• A. $-27$
• B. $36$
• C. $-36$
• D. $27$

Answer 1

The correct option is A $-27$

$x^2+y^2=9$

centre $\left(c_1\right)=\left(0,0\right)$

radius $\left(r_1\right)=3$

$x^2+y^2+6y+c=0$

centre $\left(c_2\right)=\left(0,-3\right)$

radius $\left(r_2\right)=\sqrt{g^2+f^2-c}$

$=\sqrt{0+9-c}$

$=\sqrt{9-c}$

since the circle intersect each other

So, $c_1{c}_2=r_1+r_2$

$\Rightarrow 3=3+\sqrt{9-c}$

$\Rightarrow \sqrt{9-c}=0$

$\Rightarrow c=9$

or $c_1{c}_2=r_1-r_2$

$\Rightarrow 3=3-\sqrt{9-c}$

$\Rightarrow \sqrt{9-c}=0$

$\Rightarrow c=9$

or $c_1{c}_2=r_2-r_1$

$\Rightarrow 3=\sqrt{9-c}-3$

$\Rightarrow \sqrt{9-c}=6$

$\Rightarrow 9-c=36$

$\Rightarrow c=-27$

Therefore $c=-27$