Question

How do you find the equation of the circle which has it centre at point (2,3) and passes through the origin?

Answer 1

`(x-2)^2+(y-3)^2=13`

Explanation:

If the center is at `(2,3)` and the circle passes through `(0,0)`
then the radius is `r=sqrt(2^2+3^2)=sqrt(13)`

The general equation of a circle is
`color(white)("XXX")(x-c_x)^2+(y-c_y)^2= r^2`
where `(c_x,c_y)` is the coordinate center of the circle.

graph{(x-2)^2+(y-3)^2=13 [-6.875, 13.125, -2.88, 7.12]}