Question

The tangent at $A(-1,2)$ on the circle $x^{2}+y^{2}-4 x-8 y+7=0$ touches the circle $x^{2}+y^{2}+4 x+6 y=0$ at $B .$ Then, a point of trisection of $A B$ is

• A. $\left(0, \frac{1}{3}\right)$
• B. $\left(-\frac{1}{3}, 1\right)$
• C. $\left(\frac{2}{3}, \frac{1}{3}\right)$
• D. $(-1,-1)$

Answer 1

Answer: (B)

Let circle, $C_{1}=x^{2}+y^{2}-4 x-8 y+7=0$
Centre $(2,4)$
Radius $r=\sqrt{13}$
Circle $C_{2}=x^{2}+y^{2}+4 x+6 y=0$
Centre: $(-2,-3)$

Radius $r=\sqrt{13}$
Radius of circle $C_{1}$ and $C_{2}$ are equal coordinate of $B$ is mid-point of centre of circle $C_{1}$ and $C_{2}$.
$\because B\left(0, \frac{1}{2}\right).$
$A=-1,2$
Trisection of $A B$ is
$x=\frac{2(0)+(1)(-1)}{2+1}=-\frac{1}{3}$
$y=\frac{2\left(\frac{1}{2}\right)+2( l )}{2+ l }=\frac{3}{3}=1$
$\therefore $ Coordinates are $\left(-\frac{1}{3}, 1\right)$.