Let C be the centre and A be one end of a dia | vedclass

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(D) The given circle is $x^2+y^2-2x-4y-20=0$. The centre $C$ is $(1, 2)$ and the radius $r$ is $\sqrt{1^2+2^2-(-20)} = \sqrt{25} = 5$. Since $A$ is an

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$4x^2+4y^2-8x-16y-605=0$

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(D) The given circle is $x^2+y^2-2x-4y-20=0$. The centre $C$ is $(1, 2)$ and the radius $r$ is $\sqrt{1^2+2^2-(-20)} = \sqrt{25} = 5$. Since $A$ is an end of a diameter,$A$ lies on the circle. Let $P = (h, k)$. Given that $A$ divides $CP$ in the ratio $2:3$,where $C=(1, 2)$,by the section formula,the coordinates of $A$ are: $A = \left(\frac{2h+3(1)}{2+3}, \frac{2k+3(2)}{2+3}\right) = \left(\frac{2h+3}{5}, \frac{2k+6}{5}\right)$. Since $A$ lies on the circle $x^2+y^2-2x-4y-20=0$,we substitute the coordinates of $A$ into the circle equation: $\left(\frac{2h+3}{5}\right)^2 + \left(\frac{2k+6}{5}\right)^2 - 2\left(\frac{2h+3}{5}\right) - 4\left(\frac{2k+6}{5}\right) - 20 = 0$. Multiplying by $25$: $(2h+3)^2 + (2k+6)^2 - 10(2h+3) - 20(2k+6) - 500 = 0$. $4h^2 + 12h + 9 + 4k^2 + 24k + 36 - 20h - 30 - 40k - 120 - 500 = 0$. $4h^2 + 4k^2 - 8h - 16k - 605 = 0$. Replacing $(h, k)$ with $(x, y)$,the locus of $P$ is $4x^2+4y^2-8x-16y-605=0$.

Let $C$ be the centre and $A$ be one end of a diameter of the circle $x^2+y^2-2x-4y-20=0$. If $P$ is a point such that $A$ divides $CP$ in the ratio $2:3$,then the locus of $P$ is

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