Question

A square inscribed in the circle $x^2+y^2-2x+4y+3=0$. Its side are parallel to the coordinate axes. Then, one vertex of the square is

• A. $(1+\sqrt{2},2)$
• B. $(1-\sqrt{2},-2)$
• C. $(1,-2+\sqrt{2})$
• D. $Noneofthese$

Answer 1

The correct option is D $Noneofthese$

The center of the given circle is $(1,-2)$.

Since the sides of the square inscribed in the circle are parallel to the coordinate axes, so the x-coordinate of any vertex cannot be equal to $1$ and its y-coordinate cannot be equal to $-2$.

and option A is not lying on circle.

Hence, none of the points given in (A),(B) and (C) can be the vertex of the square.