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

  $\left(0, \frac{1}{3}\right)$
• Correct answer
  B

  $\left(-\frac{1}{3}, 1\right)$
• C

  $\left(\frac{2}{3}, \frac{1}{3}\right)$
• D

  $(-1,-1)$

Solution

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 $C_2$. $$ \begin{array}{r} \because \quad B\left(0, \frac{1}{2}\right) . \\ A=-1,2 \end{array} $$ Trisection of $A B$ is $$ \begin{gathered} x=\frac{2(0)+(1)(-1)}{2+1}=-\frac{1}{3} \\ y=\frac{2\left(\frac{1}{2}\right)+2(1)}{2+1}=\frac{3}{3}=1 \end{gathered} $$ $\therefore$ Coordinates are $\left(-\frac{1}{3}, 1\right)$.

Hence, the correct answer is option: 'B'.