Question

What is the equation of a circle in general form if the center is 2,6 passes through the point 3,1? Thanks

Answer 1

`(x-2)^2+(y-6)^2=26`

Explanation:

The general form of the equation of a circle is `(x-h)^2+(y-k)^2=r^2` where `(h,k)` are the coordinates of the center and `r` is the length of the radius.

We are given that the center is at `(2,6)` so we need to figure out the length of the radius. Since we know the circle passes through the point `(3,1)`, we know the radius is just the distance between `(2,6)` and `(3,1)`. In the equation, we end up using `r^2` instead of `r` which makes our calculations a bit easier:

`"Distance"=r=sqrt((x_2-x_1)^2+(y_2-y_1)^2)`
`r^2=(x_2-x_1)^2+(y_2-y_1)^2`
`r^2=(2-3)^2+(6-1)^2`
`r^2=26`

Now that we have everything, we just substitute our values into the equation of a circle: `(x-2)^2+(y-6)^2=26`