Question

How do you find the general form of the equation of the circle with center (3,3) and containing the point (4,4)?

Answer 1

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

Explanation:

The general form for a circle is
`color(white)("XXX")(x-color(red)(a))^2+(y-color(blue)(b))=color(green)(r^2)`
where the center of the circle is at `(color(red)(a),color(blue)(b))`
and the radius is `color(green)(r)`

If the center is at `(color(red)(3),color(blue)(3))`
and `(color(brown)(4),color(orange)(4))` is a point on the circle,
then the radius is the distance from the center to (any) point on the circle
`color(white)("XXX")color(green)(r)=sqrt((color(red)(3)-color(brown)(4))^2+(color(blue)(3)-color(orange)(4)^2)`

`color(white)("XXXX")=sqrt(2)color(white)("XX")rarrcolor(white)("XX")color(green)(r^2)=color(green)(2)`

Therefore the equation of the circle is
`color(white)("XXX")(x-color(red)(3))^2+(y-color(blue)(3))^2=color(green)(2)`