AP EAMCET · Maths · Circle
The circle \(x^2+y^2-4 x-8 y+16=0\) rolls up along the tangent drawn to it at \((2+\sqrt{3}, 3)\) by 2 units. The equation of the circle in the new position is
- A \(x^2+y^2-6 x-2(4+\sqrt{3}) y+(24+8 \sqrt{3})=0\)
- B \(x^2+y^2-6 x+2(4+\sqrt{3}) y+(24+8 \sqrt{3})=0\)
- C \(x^2+y^2+6 x-2(4+\sqrt{3}) y+(24+8 \sqrt{3})=0\)
- D \(x^2+y^2+6 x+2(4+\sqrt{3}) y+(24+8 \sqrt{3})=0\)
Answer & Solution
Correct Answer
(A) \(x^2+y^2-6 x-2(4+\sqrt{3}) y+(24+8 \sqrt{3})=0\)
Step-by-step Solution
Detailed explanation
\(x^2+y^2-4 x-8 y+16=0\) \(C \equiv(2,4)\) \(\begin{aligned} & m_{\mathrm{CP}}=\frac{4-3}{2-2-\sqrt{3}}=-\frac{1}{\sqrt{3}} \\ & m_{\mathrm{CP}} \times m_{\mathrm{CC}^{\prime}}=-1\end{aligned}\)…
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