Question

The point (-1,8) is on a circle whose center is (2,6). What is the radius, equation, and what are the coordinates of the maximum point on the circle?

Answer 1

The equation of the circle is `(x-2)^2+(y-6)^2=13`, the radius is `sqrt13` (or 3.6), and the coordinates of the maximum point are`(2, 6+sqrt13)` (or `(2, 9.6))`.

Explanation:

To find the radius of the circle, we simply need to find the distance from the center to the point we are given, since we know that point is on the circle, and the radius is simply the distance from the center to the edge of the circle.

`r=sqrt((x_2-x_1)^2+(y_2-y_1)^2)`
`=sqrt((2-(-1))^2+(6-8)^2)`
`=sqrt(3^2+(-2)^2)`
`=sqrt(9+4)=sqrt13`

The equation of a circle in standard form is:

`(x-a)^2+(y-b)^2=r^2`

Where `a` is the x-coordinate of the center, `b` is the y-coordinate of the center and `r` is the radius.

The equation of this circle in standard form is:

`(x-2)^2+(y-6)^2=13`

The maximum point - the point with the highest y-value - will be directly above the center, so its x-coordinate will be the same: `2`.

Its y-coordinate will be the y-coordinate of the center plus one radius: `6+sqrt13=9.6`.

That means the coordinates of the maximum point are`(2, 6+sqrt13)` (or `(2, 9.6))`.