JEE Main · 2019 · 11 Jan 2019 · Shift 1 · Circle

A square is inscribed in the circle

x^{2} + y^{2} - 6 x + 8 y - 103 = 0

with its sides parallel to the coordinate axes. Then the distance of the vertex of the square which is nearest to the origin is:

(A)6
(B) $137$
(C) $41$ ✓ Correct
(D)13

Answer: (C)

Solution

Complete the squares in

x^{2} + y^{2} - 6 x + 8 y - 103 = 0

to get

(x - 3)^{2} + (y + 4)^{2} = 128

, a circle centred at

(3, - 4)

with radius

r = 82

. For a square inscribed with sides parallel to the axes, the diagonal is the diameter, so each vertex lies at the centre offset by

\frac{r}{2} = 8

along each axis, giving vertices

(3 \pm 8, - 4 \pm 8)

. The one nearest the origin is

(- 5, 4)

, at distance

(- 5)^{2} + 4^{2} = 41

JEE Main · 2019 · 11 Jan 2019 · Shift 1 · Single correct option

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