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A point $P$ moves so that 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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The correct answer is:
a circle with centre at $(1,4)$ and radius $\sqrt{10}$
Let point $P$ is $(h, k)$. Given $Q(0,2)$ and $R(-1,0)$ then given $P Q=\frac{1}{\sqrt{2}} P R$
Squaring on both sides,
$2(P Q)^2=(P R)^2$
$2\left(h^2+(k-2)^2\right)=(h+1)^2+k^2$
$2 h^2+2 k^2-8 k+8=h^2+1+2 h+k^2$
$\therefore \quad h^2+k^2-2 h-8 k+7=0$
$\therefore$ Locus of point $P$
$x^2+y^2-2 x-8 y+7=0$ $\ldots(i)$
$\therefore$ Eq. (i) is circle.
$\therefore$ Centre $(1,4)$ and radius $=\sqrt{1+16-7}=\sqrt{10}$
Hence, locus at point $P$ is circle with centre $(1,4)$ and radius $\sqrt{10}$
Squaring on both sides,
$2(P Q)^2=(P R)^2$
$2\left(h^2+(k-2)^2\right)=(h+1)^2+k^2$
$2 h^2+2 k^2-8 k+8=h^2+1+2 h+k^2$
$\therefore \quad h^2+k^2-2 h-8 k+7=0$
$\therefore$ Locus of point $P$
$x^2+y^2-2 x-8 y+7=0$ $\ldots(i)$
$\therefore$ Eq. (i) is circle.
$\therefore$ Centre $(1,4)$ and radius $=\sqrt{1+16-7}=\sqrt{10}$
Hence, locus at point $P$ is circle with centre $(1,4)$ and radius $\sqrt{10}$
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