Question

How do you find the equation given center passes through the origin and has its center at (0,4)?

Answer 1

`x^2+(y-4)^2=16`

Explanation:

Assuming this is to be the equation of a circle (although it was asked under the more general "Geometry of an Ellipse")

A circle with center `(color(red)a,color(blue)b)` and radius `color(green)r` has the equation:
`color(white)("XXX")(x-color(red)a)^2+(y-color(blue)b)^2=color(green)r^2`

If the required circle has its center at `(0,4)`
then `color(red)a=color(red)0`
and `color(blue)b=color(blue)4`

Further if its center is at `(0,4)` and it passes through the origin i.e. through `(0,0)` it radius must be `color(green)r=color(green)4`

So the equation of the required circle must be
`color(white)("XXX")(x-color(red)0)^2+(y-color(blue)4)^2=color(green)4^2`
or, simplifying slightly:
`color(white)("XXX")x^2+(y-4)^2=16`