A circle in the $xy$ -plane has its center at the point $(-6,1)$ . If the point $(7,12)$ lies on the circle, what is the radius of the circle?

Full solution

Q. A circle in the

xy

-plane has its center at the point

(-6,1)

. If the point

(7,12)

lies on the circle, what is the radius of the circle?

Identify Center and Point: We have: $\newline$ Center: $(-6,1)$ $\newline$ Point on the circle: $(7,12)$ $\newline$ Identify the values of the center $(h,k)$ and the point $(x,y)$ . $\newline$ $h = -6$ $\newline$ $k = 1$ $\newline$ $x = 7$ $\newline$ $y = 12$ $\newline$ The radius $r$ can be found using the distance formula between the center and the point on the circle.
Distance Formula: The distance formula is given by: $\newline$ $r = \sqrt{[(x - h)^2 + (y - k)^2]}$ $\newline$ Substitute the values of $h$ , $k$ , $x$ , and $y$ into the distance formula to find the radius. $\newline$ $r = \sqrt{[(7 - (-6))^2 + (12 - 1)^2]}$
Calculate Radius: Calculate the squares and the distance: $\newline$ $r = \sqrt{[(7 + 6)^2 + (12 - 1)^2]}$ $\newline$ $r = \sqrt{[13^2 + 11^2]}$ $\newline$ $r = \sqrt{[169 + 121]}$
Add Squares: Add the squares to find the radius: $\newline$ $r = \sqrt{169 + 121}$ $\newline$ $r = \sqrt{290}$
Find Radius: Take the square root to find the radius: $\newline$ $r = \sqrt{290}$