Question

A square is inscribed in the circle $x^2+y^2-2x+4y+3=0$, whose sides are parallel to the coordinate axes. 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

REF.Image.

Given equation of circle is $X^2+y^2-2x+4y+3=0$

$\Rightarrow (x-1{)}^2+(y+2{)}^2=2$

So, $0=Center=(1,-2)$

radius $=\sqrt{2}$

So, $AO=OC=\sqrt{2}$

Let the length of the side be a.

then length of diagnal = 2r

$\Rightarrow \sqrt{2}a=2\times \sqrt{2}$

$\Rightarrow a=2$

slope is AC is ${45}^{\circ }$ (assuming it to be lie 1 to x-axis)

So, coordinates of A, C are

$(1\pm \sqrt{2}cos{45}^{\circ },-2\pm \sqrt{2}sin{45}^{\circ })$

= (2, -1) and (0, -3)

$A=(0,-3)\Rightarrow B=(0+2,-3)=(2,-3)$

$C=(0,-3+2)=(0,-1)$

$\therefore$ Vectors are (0,-3), *(2,-3), (2,-1), (0,-1).$