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