Question

How do you write an equation of a circle with center (3,0) passes through the point (-2,4)?

Answer 1

`x^2+y^2-6x-32=0`

Explanation:

WE know that the equation of a circle with center `(a,b)` and radius `r` is `(x-a)^2-(y-b)^2=r^2`.

Here although center is given as `(3,0)`, its radius has not been given explicitly. But as circle passes through `(-2,4)` and radius is the distance between center and any point on the circumference, the radius is equal to distance between `(3,0)` and `(-2,4)`, which is

`sqrt((3-(-2))^2-(0-4)^2)=sqrt((3+2)^2-(-4)^2)=sqrt(25+16)=sqrt41`

Hence equation of circle is

`(x-3)^2-(y-0)^2=(sqrt41)^2`

or `x^2-6x+9+y^2=41`

or `x^2+y^2-6x-32=0`
graph{x^2+y^2-6x-32=0 [-16, 16, -8, 8]}