TS EAMCET · Maths · Circle
If the circle \(S=x^2+y^2+2 g x+2 f y+c=0\) cuts each of the three circles \(x^2+y^2+4 x+4 y+7=0, x^2+y^2-4 x+\) \(4 y+7=0\) and \(x^2+y^2-4 x-4 y+7=0\) orthogonally, then the equation of the tangent drawn at the point \((\sqrt{3}, 2)\) to the circle \(\mathrm{S}=0\) is
- A \((\sqrt{3}-1) x+4 y+(\sqrt{3}-1)=0\)
- B \(\sqrt{3} \mathrm{x}+2 \mathrm{y}-7=0\).
- C \((\sqrt{3}+2) \mathrm{x}+3 \mathrm{y}+(\sqrt{3}+1)=0\)
- D \(\sqrt{3} x-2 y+7=0\)
Answer & Solution
Correct Answer
(B) \(\sqrt{3} \mathrm{x}+2 \mathrm{y}-7=0\).
Step-by-step Solution
Detailed explanation
\begin{aligned} & \text {} 2 g(2)+2 f(2)=C+7 \\ & \Rightarrow \quad 4 g+4 f=C+7 \\ & 2 g(-2)+2 f(2)=C+7 \\ & \Rightarrow-4 g+4 f=C+7 \\ & 2 g(-2)+2 f(-2)=C+7 \\ & \Rightarrow \quad-4 g-4 f=C+7 \\ & (1)-(2), 8 g=0 \Rightarrow g=0 \\ & (2)+(3), C=-7 \\ & \therefore \quad f=0 \\ &…
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