AP EAMCET202315 May 2023Morning ShiftMathematicsCircleActual
The equation of the pair of tangents drawn from the point (1,1) to the circle x^2+y^2+2 x+2 y+1=0 is
Options
- A3 x^2-8 x y+3 y^2-2 x-2 y+6=0
- B11 x^2-8 x y+11 y^2-4 x-4 y-6=0
- C3 x^2-8 x y+3 y^2+2 x+2 y-2=0
- Dx^2-4 x y+y^2+x+y=0
Correct answer
C. 3 x^2-8 x y+3 y^2+2 x+2 y-2=0
Step-by-step solution
Equation of pair of tangents from point (1,1) on circle S=x^2+y^2+2 x+2 y+1 can be given as, aligned & S S₁=T^2 & (x^2+y^2+2 x+2 y+1 ) (1^2+1^2+2+2+1 ) aligned aligned & =[x(1)+y(1)+(x+1)+(y+1)+1]^2 & 7 x^2+7 y^2+14 x+14 y+7=(2 x+2 y+3)^2 & 3 x^2+3 y^2+2 x+2 y-2-8 x y=0 aligned