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

Answer: (D)

$x^2+y^2+4x-6y-12=0$
Equation of tangent at $(1,-1)$
$x-y+2(x+1)-3(y-1)-12=0$
$3x-4y-7=0$
$\therefore$ Equation of circle is
$(x^2+y^2+4x-6y-12)+\lambda (3x-4y-7)=0$
It passes through $(4,0)$ :
$(16+16-12)+\lambda (12-7)=0$
$\Rightarrow 20+\lambda (5)=0$
$\Rightarrow \lambda =-4$
$\therefore (x^2+y^2+4x-6y-12)-4(3x-4y-7)=0$
or $x^2+y^2-8x+10y+16=0$
Radius = $\sqrt{16+25-16}=5$