JEE Main · 2019 · 12 Jan 2019 · Shift 2 · Circle

If a circle of radius R passes through the origin O and intersects the coordinate axes at A and

B

, then the locus of the foot of perpendicular from

O

A B

is :

(A) $(x^{2} + y^{2})^{2} = 4 R^{2} x^{2} y^{2}$
(B) $(x^{2} + y^{2})^{3} = 4 R^{2} x^{2} y^{2}$ ✓ Correct
(C) $(x^{2} + y^{2})^{2} = 4 R x^{2} y^{2}$
(D) $(x^{2} + y^{2}) (x + y) = R^{2} x y$

Answer: (B)

Solution

Let the foot of the perpendicular from

O

to the chord

A B

(h, k)

. The chord then has equation

h x + k y = h^{2} + k^{2}

, meeting the axes at

A (\frac{h ^{2} + k ^{2}}{h}, 0)

and

B (0, \frac{h ^{2} + k ^{2}}{k})

. Since

A B

is a diameter of length

2 R

A B^{2} = 4 R^{2}

gives

\frac{( h ^{2} + k ^{2} ) ^{3}}{h ^{2} k ^{2}} = 4 R^{2}

. Replacing

(h, k)

(x, y)

, the locus is

(x^{2} + y^{2})^{3} = 4 R^{2} x^{2} y^{2}

JEE Main · 2019 · 12 Jan 2019 · Shift 2 · Single correct option

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