Question

What is the equation, center, and radius of the circle that passes through `(4,3), (0,-5),` and `(1,2)`?

Answer 1

`(x-4)^2+(y+2)^2=25`

Radius is `5`

Center is `(4,-2)`

Explanation:

A equation for a circle takes the general form

`x^2+y^2+Ax+By+C=0`

We are given `3` points that are on the circle so we can write
an equation for each point

`4^2+3^2+4A+3B+C=0`

`0^2+(-5)^2+0A+(-5)B+C=0`

`1^2+2^2+1A+2B+C=0`

Now we simplify where we can

`16+9+4A+3B+C=0`

`25-5B+C=0`

`5+1A+2B+C=0`

Now combine like terms

`25+4A+3B+C=0`

`25-5B+C=0`

`5+1A+2B+C=0`

Now rewrite by subtracting the constant term from both sides.

`4A+3B+C=-25`

`-5B+C=-25`

`1A+2B+C=-5`

I will assume that you know how to solve a system of equations

Doing so you get

`A=-8`

`B=4`

`C=-5`

So our general form for the equation of the circle is

`x^2+y^2-8x+4y-5=0`

Add `5` to both sides and rewrite

`x^2-8x+y^2+4y=5`

Now complete the squares as follows

`x^2-8x+16+y^2+4y+4=5+16+4`

`(x-4)^2+(y+2)^2=25`