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

Answer 1

Answer: (B)

Given:

The equation of the circle is $x^2+y^2-6x+8y-103=0$

Lets compare with $a{x}^2+b{y}^2+2gxx+2fy+c=0$

Here, $2g=-6\Rightarrow g=-3$

$2f=8\Rightarrow f=4$

Now, the centre of the circle $=(-g,-f)$
Centre is $(3,-4)$
Radius $=\sqrt{g^2+f^2-c}$

$=\sqrt{9+16+103}$

$=\sqrt{128}$

$=8\sqrt{2}$

Diagonal of square$=16\sqrt{2}$

Since, in a square,

$d=a\sqrt{2}$ [$a$ is side of the square]

$\Rightarrow a\sqrt{2}=16\sqrt{2}$

$\Rightarrow a=16$
Side of square $=16$
Since, the sides are parallel to coordinate axes.

and centre of the circle $=(3,-4)$

So, the coordinates of vertices of square are,

$A=(3-8,-4-8)=(-5,-12)$

$B=(3+8,-4-8)=(11,-12)$

$C=(3+8,8-4)=(11,4)$

$D=(3-8,8-4)=(-5,4)$
Therefore, the coordinates of vertices are
$A(-5,-12),B(11,-12),C(11,4),D(-5,4)$

Now,

Distance from origin to vertex $A=\sqrt{(-5{)}^2+(-12{)}^2}=\sqrt{25+144}=\sqrt{169}=13$

Distance from origin to vertex $B=\sqrt{(11{)}^2+(-12{)}^2}=\sqrt{121+144}=\sqrt{265}$

Distance from origin to vertex $C=\sqrt{(11{)}^2+(4{)}^2}=\sqrt{121+16}=\sqrt{137}$

Distance from origin to vertex $D=\sqrt{(-5{)}^2+(4{)}^2}=\sqrt{25+16}=\sqrt{41}$

Since, $\sqrt{41}<\sqrt{137}<13<\sqrt{265}$
Shortest Distance $=\sqrt{41}$