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AP EAMCET · Maths · Circle

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

  1. A \(\left(\frac{5}{2}, \frac{3}{2}\right)\)
  2. B \(\left(\frac{6}{5}, \frac{1}{5}\right)\)
  3. C \((0,-1)\)
  4. D \(\left(\frac{12}{5}, \frac{7}{5}\right)\)
Verified Solution

Answer & Solution

Correct Answer

(C) \((0,-1)\)

Step-by-step Solution

Detailed explanation

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, \(x^2+y^2-6 x-4 y+4=0\)…