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 $\mathrm{C}$ is:
Correct Option: , 3
The equation of circle $x^{2}+y^{2}+4 x-6 y=12$
can be written as $(x+2)^{2}+(y-3)^{2}=25$
Let $P=(1,-1) \& Q=(4,0)$
Equation of tangent at $P(1,-1)$ to the given circle :
$x(1)+y(-1)+2(x+1)-3(y-1)-12=0$
$3 x-4 y-7=0$$\ldots(1)$
The required circle is tangent to $(1)$ at $(1,-1)$.
$\therefore(x-1)^{2}+(y+1)^{2}+\lambda(3 x-4 y-7)=0$$\ldots(2)$
Equation (2) passes through $Q(4,0)$
$\Rightarrow 3^{2}+1^{2}+\lambda(12-7)=0 \Rightarrow 5 \lambda+10=0$
$\Rightarrow \lambda=-2$
Equation (2) becomes $x^{2}+y^{2}-8 x+10 y+16=0$
radius $=\sqrt{(-4)^{2}+(5)^{2}-16}=5$
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