TS EAMCET · Maths · Circle
The equation of the circle which touches the circle \(\mathrm{S} \equiv x^2+y^2-10 x-4 y+19=0\) at the point \((2,3)\) internally and having radius equal to half of the radius of the circle \(\mathrm{S}=0\) is
- A \(x^2+y^2+7 x+5 y+64=0\)
- B \(x^2+y^2-7 x-5 y+16=0\)
- C \(x^2+y^2-14 x-10 y+16=0\)
- D \(x^2+y^2-5 x-7 y+16=0\)
Answer & Solution
Correct Answer
(B) \(x^2+y^2-7 x-5 y+16=0\)
Step-by-step Solution
Detailed explanation
\(C_1 = (5, 2)\) \(R_1 = \sqrt{5^2+2^2-19} = \sqrt{10}\) \(R_2 = \frac{1}{2} R_1 = \frac{\sqrt{10}}{2}\) \(C_2\) divides \(C_1P\) in ratio \(R_2 : R_2 = 1:1\), so \(C_2\) is the midpoint of \(C_1P\).…
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