Q42 / 90 · JEE Main 2021 (26 Feb Shift 2) · Mathematics

If the locus of the mid-point of the line segment from the point $(3, 2)$ to a point on the circle, $x^{2} + y^{2} = 1$ is a circle of radius $r$ , then $r$ is equal to

Solution

Given circle $x^{2} + y^{2} = 1$

General coordinate on a given circle is $(\cos θ, \sin θ) .$

Given, $h = \frac{\cos θ + 3}{2} \Rightarrow \cos θ = 2 (h - \frac{3}{2}) . . . (i)$

$k = \frac{\sin θ + 2}{2} \Rightarrow \sin θ = 2 (k - 1) . . . (i i)$

Square $(i) & (i i)$ then add we get,

${(h - \frac{3}{2})}^{2} + {(k - 1)}^{2} = {(\frac{1}{2})}^{2}$

$\Rightarrow r = \frac{1}{2}$

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