If ( $h, k$ ) is the internal centre of similitude of the circles $x^2+y^2+2 x-6 y+1=0$ and $x^2+y^2-4 x+2 y+4=0$, then $4 h=$

• A

  0
• B

  3
• C

  1
• Correct answer
  D

  5

Solution

Centre and zadius of given circle $C_1(-1,3), r_1=3, C_2(2,-1), r_2=1$
$\begin{aligned} & \therefore A(k, k)=\left(\frac{6-1}{3+1}, \frac{-3+3}{3+1}\right)=\left(\frac{5}{4}, 0\right) \\ & \therefore h=\frac{5}{4} \Rightarrow 4 h=5 .\end{aligned}$

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