AP EAMCET · Maths · Circle
Let \(\mathrm{C}\) be the centre and \(\mathrm{A}\) be one end of a diameter of the circle \(x^2+y^2-2 x-4 y-20=0\). If \(P\) is a point on \(A C\) such that \(\mathrm{A}\) divides \(\mathrm{CP}\) in the ratio \(2: 3\), then the locus of \(\mathrm{P}\) is
- A \(x^2+y^2-2 x-4 y-205=0\)
- B \(2 x^2+2 y^2-4 x-8 y-405=0\)
- C \(x^2+y^2-2 x-4 y-450=0\)
- D \(4 x^2+4 y^2-8 x-16 y-605=0\)
Answer & Solution
Correct Answer
(D) \(4 x^2+4 y^2-8 x-16 y-605=0\)
Step-by-step Solution
Detailed explanation
The centre of cirlce \(x^2+y^2-2 x-4 y-20=0\) is \(\mathrm{C}(1,2)\) A.T. Q \[ \mathrm{A}(\mathrm{x}, \mathrm{y})=\left(\frac{2 \mathrm{~h}+3}{5}, \frac{2 \mathrm{k}+6}{5}\right) \] Putting the value of \(x\) and \(y\) in equation (i)…
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