AP EAMCET202316 May 2023Morning ShiftMathematicsCircleActual
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
Options
- A3 x^2+2 x y+3 y^2+8 x-8 y-16=0
- Bx^2-2 x y+y^2-4 x+4 y+8=0
- C3 x^2-2 x y-3 y^2+4 x+4 y+16=0
- Dx^2+y^2-4 x+4 y-8=0
Correct answer
A. 3 x^2+2 x y+3 y^2+8 x-8 y-16=0
Step-by-step solution
C₁: x^2+y^2-2 x+2 y-2=0C (1,-1), r= 1^2+(-1)^2+2 =2 (x-1)^2+(y+1)^2=4 ...(i) C₂: x^2+y^2=4 ...(ii) C (0,0), r=2 If P (x₁, y₁ ) i.e. pole lies outside of the circle then Chord of circle = Locus of polar (i.e. Q ) x x₁+y y₁-4=0 ...(iii) This is the equation of polar. Perpendicular distance of point O from straight line Q S=2= |1 x₁-1 y₁-4 | x₁^2+y₁^2 Squaring both sides, we get : aligned & 4 (x₁^2+y₁^2 )= (x₁-y₁-4 )^2 & 4 x₁^2+4 y₁^2=x₁^2+y₁^2+16-2 x₁ y₁+8 y₁-8 x₁ & 3 x₁^2+3 y₁^2+2 x₁ y₁+8 x₁-8 y₁-16=0 aligned Taking