A square is inscribed in the circle $x^{2} + y^{2} - 2 x + 4 y - 3 = 0$ with its sides parallel to the coordinate axes. One vertex of square is:

(3, 4)

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(3, - 4)

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(8, - 5)

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(- 8, 5)

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Solution

The correct option is A $(3, - 4)$
From equation we get
${(x - 1)}^{2} + {(y + 2)}^{2} = 8$
So centre $= (1, - 2)$ ,radius $= \sqrt{8}$
$= 2 \sqrt{2}$
So diameter $= 4 \sqrt{2}$
Now side $A B = \frac{d i a m e t e r}{\sqrt{2}} = 4$
Let coordinator of B be $(a, b)$
Since $A B ∥ x - a x i s & B C ∥ y - a x i s & A D ∥ y - a x i s$
$\Rightarrow A = (a - 4, b), C = (a, b - 4), D = (a - 4, b - 4)$
Now O is centre Hence midpoint of BD
So $1 = \frac{a + a - 4}{2}, - 2 = \frac{b + b - 4}{2}$
$6 = 2 a, - 4 = 2 b - 4$
$\Rightarrow a = 3, b = 0$
Co ordinates
$A = (- 1, 0)$
$B = (3, 0)$
$C = (3, - 4)$
$D = (- 1, - 4)$

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