Question

If square is inscribed in the circle $x^2+y^2-2x+4y+3=0$ and its sides 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. None of these

Answer 1

The correct option is D None of these

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

The sides of the square are parallel to coordinate axis

So the points where square touches the circle is $(1,-2+\sqrt{2}),(1,-2-\sqrt{2}),(1+\sqrt{2},-2),(1-\sqrt{2},-2)$

These points are midpoints of the sides of square

So the vertices of the square are $(1+\sqrt{2},-2+\sqrt{2}),(1+\sqrt{2},-2-\sqrt{2}),(1-\sqrt{2},-2+\sqrt{2}),(1-\sqrt{2},-2-\sqrt{2})$

Therefore the correct option is $D$