Question

Identify the equation of the circle that has its center at (9, 12) and passes through the origin.

Answer 1

The circle has center at (9, 12) and passes through the origin.

Equation of the circle in center and radius form is given by

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

where r is the radius and center at (h,k)

Now substitute the value of the center we get

$(x-9)^2+(y-12)^2=r^2$

As it passes through the origin so we can write

$(0-9)^2+(0-12)^2=r^2\\ \\ 81+144=r^2\\ \\ 225=r^2\\ \\ r=15$

Hence the equation of the circle is

$(x-9)^2+(y-12)^2=15^2$