Q82 / 160 · TS EAMCET 2020 (11 Sep Shift 2) · Mathematics

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

  • Correct answer
    A
    a circle with centre at $(1,4)$ and radius $\sqrt{10}$
  • B
    a parabola with focus at $(1,4)$ and length of latus rectum 10
  • C
    an ellipse with centre at $(-1,-4)$ and length of the major axis $\sqrt{10}$
  • D
    a hyperbola with centre at $(-1,-4)$ and length of the transverse axis 10

Solution

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 $\sqrt{10}$

Hence, the correct answer is option: 'A'.

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