Let's identify the equation of the circle given that its center is at [tex]\((-27,120)\)[/tex] and it passes through the origin [tex]\((0,0)\)[/tex].
First, recall the standard form of the equation of a circle with center [tex]\((h, k)\)[/tex] and radius [tex]\(r\)[/tex]:
[tex]\[
(x - h)^2 + (y - k)^2 = r^2
\][/tex]
Given:
- Center, [tex]\((h, k) = (-27, 120)\)[/tex]
- A point on the circle, [tex]\((x_1, y_1) = (0, 0)\)[/tex]
Next, we need to calculate the radius [tex]\(r\)[/tex], which is the distance from the center of the circle to the origin. We can use the distance formula for this, which is:
[tex]\[
r = \sqrt{(x_1 - h)^2 + (y_1 - k)^2}
\][/tex]
Substitute the center [tex]\((h, k) = (-27, 120)\)[/tex] and the point [tex]\((x_1, y_1) = (0, 0)\)[/tex]:
[tex]\[
r = \sqrt{(0 + 27)^2 + (0 - 120)^2} = \sqrt{27^2 + (-120)^2}
\][/tex]
Calculating [tex]\(27^2\)[/tex] and [tex]\((-120)^2\)[/tex]:
[tex]\[
27^2 = 729
\][/tex]
[tex]\[
(-120)^2 = 14400
\][/tex]
Now, add these values to find [tex]\(r^2\)[/tex]:
[tex]\[
r^2 = 729 + 14400 = 15129
\][/tex]
Thus, the radius-squared [tex]\(r^2\)[/tex] is 15129.
Now, substitute the center [tex]\((-27, 120)\)[/tex] and [tex]\(r^2 = 15129\)[/tex] into the standard form of the circle's equation:
[tex]\[
(x - (-27))^2 + (y - 120)^2 = 15129
\][/tex]
Simplify the equation:
[tex]\[
(x + 27)^2 + (y - 120)^2 = 15129
\][/tex]
Thus, the equation of the circle is:
[tex]\[
(x + 27)^2 + (y - 120)^2 = 15129
\][/tex]
Therefore, the correct option is:
D. [tex]\((x+27)^2+(y-120)^2=15129\)[/tex]
