AP EAMCET · Maths · Circle
The point of intersection of the tangents drawn at the points where the line \(2 x-y+3=0\) meets the circle \(x^2+\) \(y^2-4 x-6 y+4=0\) is
- A \(\left(-8, \frac{15}{2}\right)\)
- B \(\left(\frac{-5}{2}, \frac{21}{4}\right)\)
- C \(\left(\frac{5}{2}, \frac{-21}{4}\right)\)
- D \(\left(8, \frac{-15}{2}\right)\)
Answer & Solution
Correct Answer
(B) \(\left(\frac{-5}{2}, \frac{21}{4}\right)\)
Step-by-step Solution
Detailed explanation
Given: \(x^2+y^2-4 x-6 y+4=0\) \(\Rightarrow(x-2)^2+(y-3)^2=3^2\) The given equation of line is \(2 x-y+3=0\) The required point is…
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