Question

By using the equation below complete the square, write the equation of the circle in standard form from the following equation. Then determine the center and radius of
a circle.

10y + x2 = -18+ 5x - y2

Answer 1

Given:

The equation of a circle is

$10y+x^2=-18+5x-y^2$

To find:

The center and radius of the given equation by completing the square.

Solution:

The standard form of a circle is

$(x-h)^2+(y-k)^2=r^2$            ...(i)

where, (h,k) is center and r is radius of the circle.

We have,

$10y+x^2=-18+5x-y^2$

It can be written as

$(x^2-5x)+(y^2+10y)=-18$

$\left(x^2-5x+\left(\dfrac{5}{2}\right)^2\right)+\left(y^2+10y+\left(\dfrac{10}{2}\right)^2\right)=-18+\left(\dfrac{5}{2}\right)^2+\left(\dfrac{10}{2}\right)^2$

$\left(x-\dfrac{5}{2}\right)^2+\left(y^2+10y+5^2\right)=-18+\dfrac{25}{4}+5^2$

$\left(x-\dfrac{5}{2}\right)^2+(y+5)^2=-18+\dfrac{25}{4}+25$

$\left(x-\dfrac{5}{2}\right)^2+(y+5)^2=\dfrac{-72+25+100}{4}$

$\left(x-\dfrac{5}{2}\right)^2+(y+5)^2=\dfrac{53}{4}$

$\left(x-\dfrac{5}{2}\right)^2+(y+5)^2=\left(\dfrac{\sqrt{53}}{2}\right)^2$          ...(ii)

On comparing (i) and (ii), we get

$h=\dfrac{5}{2},k=-5,r=\dfrac{\sqrt{53}}{2}$

Therefore, the center is $\left(\dfrac{5}{2},-5\right)$ and the radius is $\dfrac{\sqrt{53}}{2}$ units.