Question

A square is inscribed in the circle $x^2+y^2-2x+4y-93=0$ with its sides parallel to the axes of coordinates. The coordinates of the vertex farthest from the origin are

• A. $(9,-8)$
• B. $(8,-9)$
• C. $(8,5)$
• D. $(-6,9)$

Answer 1

The correct option is B $(8,-9)$

The center of the given circle is $(1,-2)$ and its radius is $\sqrt{98}$,

so the coordinate of the vertices of the square inscribed in the given circle are $(1+\sqrt{98}\cos \theta ,-2+\sqrt{98}\sin \theta )$ where $\theta =\pm \frac{\pi }{4},\pm \frac{3\pi }{4}$

$S^2$ square of the distance of a vertex from the origin

$=5+98+2\sqrt{98}(\cos \theta -2\sin \theta )$

which is maximum when $\theta =-\frac{\pi }{4}$.

So the required point is $(1+7,-2-7)=(8,-9)$