Question

Identify the equation of the circle that has its center at (16, 30) and passes through the origin

Answer 1

To solve this question, we have to find the equation of the circle with given center and where it passes. Doing this, we get that the equation of the circle is:

$(x - 16)^2 + (y - 30)^2 = 1156$

Equation of a circle:

The equation of a circle with center $(x_0, y_0)$ and radius r is given by:

$(x - x_0)^2 + (y - y_0)^2 = r^2$

Center at (16, 30)

This means that $x_0 = 16, y_0 = 30$

Thus

$(x - 16)^2 + (y - 30)^2 = r^2$

Passes through the origin:

We use this to find the radius squared, as this means that $x = 0, y = 0$ is part of the circle. Thus

$(x - 16)^2 + (y - 30)^2 = r^2$

$(0 - 16)^2 + (0 - 30)^2 = r^2$

$r^2 = 16^2 + 30^2 = 1156$

Thus, the equation of the circle is:

$(x - 16)^2 + (y - 30)^2 = 1156$

For another example to find the equation of a circle, you can look at brainly.com/question/23719612