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
- A \(\left(\frac{5}{2}, \frac{3}{2}\right)\)
- B \(\left(\frac{6}{5}, \frac{1}{5}\right)\)
- C \((0,-1)\)
- D \(\left(\frac{12}{5}, \frac{7}{5}\right)\)
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, \(x^2+y^2-6 x-4 y+4=0\)…
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