JEE Main · 2021 · 26 Feb 2021 · Shift 2 · Circle

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 :

(A) $\frac{1}{3}$
(B) $\frac{1}{4}$
(C) $\frac{1}{2}$ ✓ Correct
(D)1

Answer: (C)

Solution

Let

(h, k)

be the mid-point of the segment joining

(3, 2)

to the point

(cos θ, sin θ)

of the circle. Then

h = \frac{3 + cos θ}{2}, k = \frac{2 + sin θ}{2}

cos θ = 2 h - 3

and

sin θ = 2 k - 2

. Using

cos^{2} θ + sin^{2} θ = 1

(2 h - 3)^{2} + (2 k - 2)^{2} = 1 ⟺ (h - \frac{3}{2})^{2} + (k - 1)^{2} = \frac{1}{4}

The locus is a circle of centre

(\frac{3}{2}, 1)

, so

r = \frac{1}{2}

JEE Main · 2021 · 26 Feb 2021 · Shift 2 · Single correct option

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