Question

Circle O with center (x,y), passes through the points A(0,0), B(-3,0), and C(1, 2). Find the coordinates of the center of the circle

In your final answer, include all formulas and calculations used to find Point O, x,y). the center of circle O.

Answer 1

The center has the coordinates
`(-1.5, 2)` and the radius 2.5

Explanation:

Let
`(x_0,y_0)` be the center of the circel and r be the radius, then the equation of the circle is

`(x-x_0)^2+(y-y_0)^2=r^2`

Circle O with center
`(x_0,y_0)` passes through the points A(0,0), B(-3,0), and C(1, 2), so

`\begin{array}{l}(0-x_0)^2+(0-y_0)^2=r^2\\ \\(-3-x_0)^2+(0-y_0)^2=r^2\\ \\(1-x_0)^2+(2-y_0)^2=r^2\end{array}\Rightarrow \begin{array}{l}x_0^2+y_0^2=r^2\\ \\(3+x_0)^2+y_0^2=r^2\\ \\(1-x_0)^2+(2-y_0)^2=r^2\end{array}`

Subtract from the second equation the first one:

`(3+x_0)^2+y_0^2-x_0^2-y_0^2=r^2-r^2\\ \\(3+x_0)^2-x_0^2=0\\ \\(3+x_0)^2=x_0^2\\ \\3+x_0=x_0\ \text{or}\ 3+x_0=-x_0\\ \\3=0\ \text{or}\ 2x_0=-3,\ x_0=-1.5`

Substitute it into the last two equations:

`\begin{array}{l}(3-1.5)^2+y_0^2=r^2\\ \\(1+1.5)^2+(2-y_0)^2=r^2\end{array}\Rightarrow \begin{array}{l}2.25+y_0^2=r^2\\ \\6.25+(2-y_0)^2=r^2\end{array}`

Subtract them:

`6.25+(2-y_0)^2-2.25-y_0^2=r^2-r^2\\ \\4+4-4y_0+y_0^2-y_0^2=0\\ \\4y_0=8\\ \\y_0=2`

Substitute into the first equation:

`(-1.5)^2+2^2=r^2\\ \\r^2=2.25+4\\ \\r^2=6.25\\ \\r=2.5`

So, the center has the coordinates
`(-1.5, 2)` and the radius 2.5