Question

How do you write the equation of the circle with center(1,-2) and passes through (6,-6)?

Answer 1

`(x-1)^2+(y+2)^2 = 41`

Explanation:

The equation of a circle with centre `(h, k)` and radius `r` may be written:

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

We are given `(h,k) = (1,-2)`, so the only unknown is `r^2`.

Since the circle passes through `(6, -6)`, the values `x=6`, `y=-6` will satisfy the equation and we find:

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

`= (6-1)^2+((-6)-(-2))^2`

`= 5^2+(-4)^2`

`= 25+16`

`= 41`

So the equation of our circle may be written:

`(x-1)^2+(y+2)^2 = 41`

graph{((x-1)^2+(y+2)^2-41)((x-1)^2+(y+2)^2-0.02)((x-6)^2+(y+6)^2-0.02) = 0 [-8.58, 11.42, -8.52, 1.48]}