Question

A square is inscribed in the circle $x^2+y^2-2x+4y-93=0$ with its sides parallel to the coordinates axes. The coordinates of its vertices are

• A. $(-6,9),(-6,-5),(8,-9),(8,5)$
• B. $(-6,-9),(-6,5),(8,-9),(8,-5)$
• C. $(-6,-9),(-6,5),(8,9),(8,5)$
• D. $(-6,-9),(-6,5),(8,5),(8,-9)$

Answer 1

The correct option is D $(-6,-9),(-6,5),(8,5),(8,-9)$

The equation is of the form $x^2+y^2+2hx+2ky+c=0$ where $r=\sqrt{h^2+k^2-c}$

$\therefore h=-1,k=2,c=-93$ and $r=\sqrt{(-1{)}^2+(2{)}^2-(-93)}=\sqrt{98}=7\sqrt{2}$ and centre$O(-h,-k)=O(1,-2)$

Now suppose the square inscribed with its sides parallel to the coordinates axes is $ABCD$ and centre of circle is $O.$

So, $AB$ will be one of the chords of circle, thats why distance from centre $O$ to line $AB$ is $\frac{r}{\sqrt{2}}=7$

Now, equation of line parallel to x-axis passing through $O$ is $y=1$,

So equation of a line parallel to $O$ and at a distance $7$ from it will be $y=1+7\left(AB\right)\to y=8$ and $y=1-7\left(CD\right)\to y=-6$

Similarly, equation of line parallel to y-axis passing through $O$ is $x=-2$

$\therefore x=-2-7\left(AC\right)\to x=-9$ and $x=-2+7\left(BD\right)\to x=5$

Now, intersection of $AB$ and $AC$ gives point $A$, which will be $A(8,-9)$

Similarly, $B(8,5),C(-6,-9),D(-6,5)$

Hence, points are $(-6,-9),(-6,5),(8,5),(8,-9)$ $=\left(D\right)$