JEE Main · 2019 · 09 Apr 2019 · Shift 1 · Circle

If a tangent to the circle

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

intersects the coordinate axes at distinct points

P

and

Q

, then the locus of the mid-point of

P Q

is:

(A) $x^{2} + y^{2} - 16 x^{2} y^{2} = 0$
(B) $x^{2} + y^{2} - 2 x^{2} y^{2} = 0$
(C) $x^{2} + y^{2} - 4 x^{2} y^{2} = 0$ ✓ Correct
(D) $x^{2} + y^{2} - 2 x y = 0$

Answer: (C)

Solution

Let the midpoint of

P Q

(h, k)

, with the tangent cutting the axes at

P (2 h, 0)

and

Q (0, 2 k)

. That line is

\frac{x}{2 h} + \frac{y}{2 k} = 1

, and being tangent to

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

its distance from the origin is

1

\frac{1}{\frac{1}{4 h ^{2}} + \frac{1}{4 k ^{2}}} = 1

, so

\frac{1}{4 h ^{2}} + \frac{1}{4 k ^{2}} = 1

. Clearing denominators,

k^{2} + h^{2} = 4 h^{2} k^{2}

, so the locus is

x^{2} + y^{2} - 4 x^{2} y^{2} = 0

JEE Main · 2019 · 09 Apr 2019 · Shift 1 · Single correct option

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