TS EAMCET · Maths · Circle
A line meets the circle \(x^2+y^2-4 x-4 y-8=0\) in two points A and B. If \(\mathrm{P}(2,-2)\) is a point on the circle such that \(\mathrm{PA}=\mathrm{PB}=2\) then the equation of the line AB is
- A \(2 x+3 y=0\)
- B \(3 x+2 y=0\)
- C \(2 x+3=0\)
- D \(2 y+3=0\)
Answer & Solution
Correct Answer
(D) \(2 y+3=0\)
Step-by-step Solution
Detailed explanation
\(S_1: x^2+y^2-4x-4y-8=0\) \(S_2: (x-2)^2+(y+2)^2=2^2 \implies x^2-4x+4+y^2+4y+4=4 \implies x^2+y^2-4x+4y+4=0\) \(\text{Line AB is } S_1-S_2=0\) \((x^2+y^2-4x-4y-8)-(x^2+y^2-4x+4y+4)=0\) \(-8y-12=0\) \(2y+3=0\)
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