The point of intersection of the common tangents drawn to the circles \(x^2+y^2-4 x-2 y+1=0\) and \(x^2+y^2-6 x-4 y+4=0\), is

• A

  \(\left(\frac{5}{2}, \frac{3}{2}\right)\)
• B

  \(\left(\frac{6}{5}, \frac{1}{5}\right)\)
• Correct answer
  C

  \((0,-1)\)
• D

  \(\left(\frac{12}{5}, \frac{7}{5}\right)\)

Solution

Given equation of circles are \(\begin{aligned} & x^2+y^2-4 x-2 y+1=0 \quad \ldots (i) \\ & \text{and } x^2+y^2-6 x-4 y+4=0 \quad \ldots (ii) \end{aligned}\) Here, \(x^2+y^2-6 x-4 y+4=0\) \(\begin{aligned} & C_1=(2,1), C_2=(3,2) \\ & r_1=\sqrt{4+1-1}=\sqrt{4}=2 \\ & \text{and } r_2=\sqrt{9+4-4}=\sqrt{9}=3 \end{aligned}\) and \(r_2=\sqrt{9+4-4}=\sqrt{9}=3\) Now, \(\begin{aligned} C_1 C_2 & =\sqrt{(3-2)^2+(2-1)^2} \\ & =\sqrt{1+1}=\sqrt{2} \end{aligned}\) \(\begin{array}{ll} \text {and } & r_1+r_2=2+3=5 \\ \therefore & C_1 C_2 < r_1+r_2 \end{array}\) So, circles intersect at two distinct point. let \(P(x, y)\) be the point of intersection of tangents \(\therefore \quad P(x, y)=\left(\frac{6-6}{-1}, \frac{4-3}{-1}\right)\)
Hence, \(P(x, y)=(0,-1)\)

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