Question

what is the equation of a circle that has a center at the origin and passes through the point (-6, 8)

Answer 1

To find the equation of a circle centered at the origin (0, 0) that passes through the point (-6, 8), we can use the standard form of the equation of a circle:

\[
x^2 + y^2 = r^2
\]

where \( r \) is the radius of the circle.

First, we need to determine the radius \( r \). The radius is the distance from the center of the circle to a point on the circle. We can calculate this distance using the point (-6, 8):

\[
r = \sqrt{(-6 - 0)^2 + (8 - 0)^2}
\]

Calculating this, we get:

\[
r = \sqrt{(-6)^2 + (8)^2} = \sqrt{36 + 64} = \sqrt{100} = 10
\]

Now, we substitute \( r^2 \) into the equation of the circle:

\[
r^2 = 10^2 = 100
\]

Therefore, the equation of the circle is:

\[
x^2 + y^2 = 100
\]

This is the equation of the circle centered at the origin and passing through the point (-6, 8).