To find the equation of a circle given a center [tex]\((h, k)\)[/tex] and a point on the circle [tex]\((x_1, y_1)\)[/tex], we need to use the standard form of the circle's equation:
[tex]\[
(x - h)^2 + (y - k)^2 = r^2
\][/tex]
where [tex]\((h, k)\)[/tex] is the center of the circle and [tex]\(r\)[/tex] is the radius.
Given:
- The center of circle [tex]\(C\)[/tex] is [tex]\((-2, 10)\)[/tex], so [tex]\(h = -2\)[/tex] and [tex]\(k = 10\)[/tex].
- The circle passes through the point [tex]\(P(10, 5)\)[/tex].
First, we calculate the radius squared [tex]\((r^2)\)[/tex]. We use the distance formula to find the distance from the center [tex]\((-2, 10)\)[/tex] to the point [tex]\(P(10, 5)\)[/tex], since this distance is the radius [tex]\(r\)[/tex] of the circle:
[tex]\[
r^2 = (x_1 - h)^2 + (y_1 - k)^2
\][/tex]
Substituting the given coordinates, we get:
[tex]\[
r^2 = (10 - (-2))^2 + (5 - 10)^2
\][/tex]
Simplify the expression:
[tex]\[
r^2 = (10 + 2)^2 + (5 - 10)^2
\][/tex]
[tex]\[
r^2 = 12^2 + (-5)^2
\][/tex]
[tex]\[
r^2 = 144 + 25
\][/tex]
[tex]\[
r^2 = 169
\][/tex]
Now, substitute [tex]\(h = -2\)[/tex], [tex]\(k = 10\)[/tex], and [tex]\(r^2 = 169\)[/tex] into the standard form of the circle's equation:
[tex]\[
(x - (-2))^2 + (y - 10)^2 = 169
\][/tex]
Simplify the equation:
[tex]\[
(x + 2)^2 + (y - 10)^2 = 169
\][/tex]
Therefore, the equation that represents circle [tex]\(C\)[/tex] is:
[tex]\[
(x + 2)^2 + (y - 10)^2 = 169
\][/tex]
The correct answer is:
D. [tex]\((x + 2)^2 + (y - 10)^2 = 169\)[/tex]
