A point P moves such that the distance from ( | vedclass

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Question

(A) Let the point $P$ be $(x, y)$. Given $Q(0,2)$ and $R(-1,0)$,the condition is $PQ = \frac{1}{\sqrt{2}} PR$. Squaring both sides,we get $2(PQ)^2 = (

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a circle with centre at $(1,4)$ and radius $\sqrt{10}$

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(A) Let the point $P$ be $(x, y)$. Given $Q(0,2)$ and $R(-1,0)$,the condition is $PQ = \frac{1}{\sqrt{2}} PR$. Squaring both sides,we get $2(PQ)^2 = (PR)^2$. Substituting the coordinates,$2(x^2 + (y-2)^2) = (x+1)^2 + y^2$. Expanding the terms: $2(x^2 + y^2 - 4y + 4) = x^2 + 2x + 1 + y^2$. $2x^2 + 2y^2 - 8y + 8 = x^2 + 2x + 1 + y^2$. Rearranging the terms: $x^2 + y^2 - 2x - 8y + 7 = 0$. This is the equation of a circle. The centre is $(-\frac{-2}{2}, -\frac{-8}{2}) = (1, 4)$. The radius is $\sqrt{g^2 + f^2 - c} = \sqrt{1^2 + 4^2 - 7} = \sqrt{1 + 16 - 7} = \sqrt{10}$. Thus,the locus is a circle with centre $(1, 4)$ and radius $\sqrt{10}$.

$A$ point $P$ moves such that the distance from $(0,2)$ to $P$ is $\frac{1}{\sqrt{2}}$ times the distance of $P$ from $(-1,0)$. Then the locus of the point is

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