JEE Main · 2019 · 10 Apr 2019 · Shift 2 · Circle

The locus of the centres of the circles, which touch the circle,

x^{2} + y^{2} = 1

externally, also touch the

y

-axis and lie in the first quadrant is

(A) $x = 1 + 2 y, y \geq 0$
(B) $y = 1 + 4 x, x \geq 0$
(C) $x = 1 + 4 y, y \geq 0$
(D) $y = 1 + 2 x, x \geq 0$ ✓ Correct

Answer: (D)

Solution

Let the moving circle have centre

(h, k)

with

h, k > 0

and radius

r

. Touching the

y

-axis fixes

r = h

. Touching

x^{2} + y^{2} = 1

externally means the centres are

1 + r

apart,

h^{2} + k^{2} = 1 + h

. Squaring,

h^{2} + k^{2} = 1 + 2 h + h^{2}

, so

k^{2} = 1 + 2 h

. Writing

(h, k)

(x, y)

, the locus is

y = 1 + 2 x

for

x \geq 0

JEE Main · 2019 · 10 Apr 2019 · Shift 2 · Single correct option

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