Question

How do you write an equation of a circle given center at the point (1, 0) and has the point (-2, 3)?

Answer 1

`x^2+y^2-2x-17=0`

Explanation:

The equation of the circle with center at (a, b) and radius r is
`(x-a)^2+(y-b)^2=r^2`.

Here a = 1 and b = 0.
The equation becomes
`(x-1)^2+y^2=r^2`.

The circle passes through (`-2`, 3).
So, `r^2` = 18.
Answer .`(x-1)^2+y^2=18` or .`x^2+y^2-2x-17=0`