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…

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
  1. $x^2+y^2-2 x-4 y-205=0$
  2. $2 x^2+2 y^2-4 x-8 y-405=0$
  3. $x^2+y^2-2 x-4 y-450=0$
  4. $4 x^2+4 y^2-8 x-16 y-605=0$

Solution

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) $ \begin{aligned} & \left(\frac{2 h+3}{5}\right)^2+\left(\frac{2 k+6}{5}\right)^2-2 \frac{(2 h+3)}{5}-4 \frac{(2 k+6)}{5}-20=0 \\ & 4 h^2+9+12 h+4 k^2+36+24 k-20 h-30-40 k-120 \\ & -500=0 \\ & 4 h^2+4 k^2-8 h-16 k-605=0 \\ & \therefore \text { Locus of } P \text { is } 4 x^2+4 y^2-8 x-16 y-605=0 \end{aligned} $

Asked in: AP EAMCET 2023 (18 May Shift 2)

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