Question

If a circle $C$ passing through the point $(4,0)$ touches the circle $x^2+y^2+4x-6y=12$ externally at the point $(1,-1)$ then the radius of $C$ is :

• A. $\sqrt{57}$
• B. $4$
• C. $2\sqrt{5}$
• D. $5$

Answer 1

The correct option is D $5$

Equation of tangent $T$, at point $(1,-1)$ is
$x\left(1\right)+y(-1)+2(x+1)-3(y-1)-12=0$
$\Rightarrow 3x-4y=7$
$\therefore$ Required circle is
$(x-1{)}^2+(y+1{)}^2+\lambda (3x-4y-7)=0\cdots \left(1\right)$
which passes through $(4,0)$
$\Rightarrow 9+1+\lambda (12-7)=0$
$\Rightarrow \lambda =-2$
From equation $\left(1\right)$
$x^2+y^2-8x+10y+16=0$
$\Rightarrow$ Radius $=\sqrt{16+25-16}=5$