Question

The tangent at $A(-1,2)$ on the circle $x^2+y^2-4x-8y+7=0$ touches the circle $x^2+y^2+4x+6y=0$ at $B.$ Then, a point of trisection of $AB$ 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, $C_1=x^2+y^2-4x-8y+7=0$
Centre $(2,4)$
Radius $r=\sqrt{13}$
Circle $C_2=x^2+y^2+4x+6y=0$
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 $AB$ 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 $\left(-\frac{1}{3},1\right)$.