AP EAMCET · Maths · Circle
The locus of the poles of the tangents to the circle \(x^2+y^2-2 x+2 y-2=0\) with respect to the circle \(x^2+y^2=4\), is
- A \(3 x^2+2 x y+3 y^2+8 x-8 y-16=0\)
- B \(x^2-2 x y+y^2-4 x+4 y+8=0\)
- C \(3 x^2-2 x y-3 y^2+4 x+4 y+16=0\)
- D \(x^2+y^2-4 x+4 y-8=0\)
Answer & Solution
Correct Answer
(A) \(3 x^2+2 x y+3 y^2+8 x-8 y-16=0\)
Step-by-step Solution
Detailed explanation
\(C_1: x^2+y^2-2 x+2 y-2=0\) \(C \equiv(1,-1), r=\sqrt{1^2+(-1)^2+2}=2\) \(\Rightarrow(x-1)^2+(y+1)^2=4\) ...(i) \(C_2: x^2+y^2=4\) ...(ii) \(C \equiv(0,0), r=2\) If \(P\left(x_1, y_1\right)\) i.e. pole lies outside of the circle then Chord of circle \(=\) Locus of polar (i.e.…
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