Question

How do you find an equation of the circle that passes through the pts (0,0) and (1,1) and has its center on the line y=2?

Answer 1

`(x+1)^2+(y-2)^2=5`

Explanation:

If the center of the circle is on `y=2`
then the center of the circle is at `(x_c,2)` for some value `x_c`

The radius of the circle, `r`, is

• The distance from `(x_c,2)` to `(0,0)`
  `color(white)("XXX")=sqrt((x_c-0)^2+(2-0)^2)=sqrt(x_c^2+4)`
• The distance from `(x_c,2)` to `(1,1)`
  `color(white)("XXX")=sqrt((x_c-1)^2+(2-1)^2) = sqrt(x_c^2-2x_c+1+1)`

Therefore
`color(white)("XXX")x_c^2+4 = x_c^2-2x_c+2`

`color(white)("XXX")-2x_c= -2`

`color(white)("XXX")x_c= -1`

and since `r=sqrt(x_c^2+4)`
`color(white)("XXX")r=sqrt(5)`

The general equation of a circle with center `(a,b)` and radius `r` is
`color(white)("XXX")(x-a)^2+(y-b)^2=r^2`

By substituting `(-1,2)` for `(a,b)` and `sqrt(5)` for `r`, we get
`color(white)("XXX")(x+1)^2+(y-2)^2=5`
graph{(x+1)^2+(y-2)^2=5 [-5.8, 5.3, -0.764, 4.783]}