A circle graphed in the $xy$ -plane has its center at $(2,3)$ . If the point $(8,11)$ lies on the circle, which of the following is an equation of the circle?

Full solution

Q. A circle graphed in the

xy

-plane has its center at

(2,3)

. If the point

(8,11)

lies on the circle, which of the following is an equation of the circle?

Find Radius: We need to find the radius of the circle using the distance formula between the center ( $2$ , $3$ ) and the point ( $8$ , $11$ ) on the circle. $\newline$ The distance formula is: $r = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$ $\newline$ Here, $(x_1, y_1) = (2, 3)$ and $(x_2, y_2) = (8, 11)$ . $\newline$ So, $r = \sqrt{(8 - 2)^2 + (11 - 3)^2}$ $\newline$ $r = \sqrt{6^2 + 8^2}$ $\newline$ $r = \sqrt{36 + 64}$ $\newline$ $r = \sqrt{100}$ $\newline$ $r = 10$
Distance Formula: Now that we have the radius, we can write the equation of the circle in standard form. $\newline$ The standard form of a circle's equation is $(x - h)^2 + (y - k)^2 = r^2$ , where $(h, k)$ is the center of the circle and $r$ is the radius. $\newline$ Here, $h = 2$ , $k = 3$ , and $r = 10$ . $\newline$ So, the equation of the circle is $(x - 2)^2 + (y - 3)^2 = 10^2$ .
Standard Form: Simplify the equation by squaring the radius. $\newline$ $(x - 2)^2 + (y - 3)^2 = 100$ $\newline$ This is the equation of the circle in standard form.