Question

A square is inscribed in the circle $x^2+y^2-6x+8y-103=0$ with its sides parallel to the coordinate axes. Then the distance of the vertex of this square which is nearest to the origin is

• A. $\sqrt{41}$
• B. $13$
• C. $\sqrt{137}$
• D. $6$

Answer 1

The correct option is A $\sqrt{41}$

Given circle : $x^2+y^2-6x+8y-103=0$
Centre and radius is,
$C\equiv (-g,-f)=(3,-4),r=\sqrt{g^2+f^2-c}\Rightarrow r=\sqrt{9+16+103}=8\sqrt{2}$


Since, sides of square are parallel to the coordinate axes, diagonal will be parallel to $y=x,y=-x$

$PR$ is parallel to $y=x$
$QS$ is parallel to $y=-x$
For $y=x,$
$\text{inclination}={45}^{\text{o}}$
For $y=-x,$
$\text{inclination}={135}^{\text{o}}$

Using parametric equations of a straight line
$R,P=(3\pm 8\sqrt{2}\cos {45}^{\circ },-4\pm 8\sqrt{2}\sin {45}^{\circ })$
$\Rightarrow R=(11,4),P=(-5,-12)S,Q=(3\pm 8\sqrt{2}\cos {135}^{\circ },-4\pm 8\sqrt{2}\sin {135}^{\circ })$
$\Rightarrow S=(-5,4),Q=(11,-12)$

Now, $OP=\sqrt{25+144}=\sqrt{169}=13,$
$OR=\sqrt{121+16}=\sqrt{137},OQ=\sqrt{121+144}=\sqrt{265},OS=\sqrt{25+16}=\sqrt{41}$
$\therefore S$ is the nearest to the origin.