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Question

A square is inscribed in the circle x2+y2−2x+4y−93=0 with its sides parallel to the coordinate axes. The coordinates of its vertices are


A

(-6, -9), (-6, 5), (8, -9) and (8, 5)

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B

(-6, 9), (-6, -5), (8, -9) and (8, 5)

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C

(-6, -9), (-6, 5), (8, 9) and (8, 5)

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D

(-6, -9), (-6, 5), (8, -9) and (8, -5)

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Solution

The correct option is A

(-6, -9), (-6, 5), (8, -9) and (8, 5)


Let x = a, x = b, y = c and y = d be the sides of the square. The length of each diagonal of the square is equal

to the diameter of the circle i.e., 21+4+93. Let l be the length of each side of the square.

Then 2l2=(Diagonal)22l2=(2.1+4+93)2 l = 14

Therefore each side of the square is at a distance 7 from the centre (1, -2) of the given circle. This implies

that a = -6, b = 8, c = -9, d = 5.

Therefore the vertices of the square are (-6, -9), (-6, 5), (8, -9), (8, 5).


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