Deepak Scored 45->99%ile with Bounce Back Crack Course. You can do it too!

# Solve this following

Question:

If a circle $C$ passing through the point $(4,0)$

touches the circle $x^{2}+y^{2}+4 x-6 y=12$ externally at the point $(1,-1)$, then the radius of $C$ is :

1. $\sqrt{57}$

2. 4

3. $2 \sqrt{5}$

4. 5

Correct Option: , 4

Solution:

$x^{2}+y^{2}+4 x-6 y-12=0$

Equation of tangent at $(1,-1)$

$x-y+2(x+1)-3(y-1)-12=0$

$3 x-4 y-7=0$

$\therefore$ Equation of circle is

$\left(x^{2}+y^{2}+4 x-6 y-12\right)+\lambda(3 x-4 y-7)=0$

It passes through $(4,0)$ :

$(16+16-12)+\lambda(12-7)=0$

$\Rightarrow 20+\lambda(5)=0$

$\Rightarrow \lambda=-4$

$\therefore\left(x^{2}+y^{2}+4 x-6 y-12\right)-4(3 x-4 y-7)=0$

or $x^{2}+y^{2}-8 x+10 y+16=0$

Radius $=\sqrt{16+25-16}=5$