Question

One diameter of the circle circumscribing the rectangle $ABCD$ is $4y=x+7$. If the coordinates of $A$ and $B$ are $(-3,4)$ and $(5,4)$ respectively, find the equation of the circle.

Answer 1

Clearly, the centre of the circle lies on the line $4y=x+7$

The circle passes through $A(-3,4)$ and $B(5,4)$

The slope of the segment joining $A$ and $B$ is zero.

$\therefore$ the slope of the perpendicular bisector of $AB$ is not defined.

Hence the perpendicular bisector of $AB$ will be parallel to the $y-$axis and will pass through $(\frac{-3+5}{2},\frac{4+4}{2})=(1,4)$

The equation of the perpendicular bisector is $x=1$

The intersection point of the perpendicular bisector and $4y=x+7$ is $(1,2)$

$\therefore$ Centre$=(1,2)$

Radius$=\sqrt{{(5-1)}^2+{(4-2)}^2}=\sqrt{20}$units.

Hence the required equation of the circle is ${(x-1)}^2+{(y-2)}^2={\left(\sqrt{20}\right)}^2$

or $x^2-2x+1+y^2-4y+4-20=0$

or $x^2+y^2-2x-4y-15=0$