AP EAMCET · Maths · Circle
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
- A \(3 x^2-8 x y+3 y^2-2 x-2 y+6=0\)
- B \(11 x^2-8 x y+11 y^2-4 x-4 y-6=0\)
- C \(3 x^2-8 x y+3 y^2+2 x+2 y-2=0\)
- D \(x^2-4 x y+y^2+x+y=0\)
Answer & Solution
Correct Answer
(C) \(3 x^2-8 x y+3 y^2+2 x+2 y-2=0\)
Step-by-step Solution
Detailed explanation
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, \[ \begin{aligned} & S S_1=T^2 \\ & \Rightarrow\left(x^2+y^2+2 x+2 y+1\right)\left(1^2+1^2+2+2+1\right) \end{aligned} \]…
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