Question

Find an equation of the circle that satisfies the given conditions. (Use the variables x and y.)

Answer 1

Answer:

$x^2 + y^2 = 52$

Step-by-step explanation:

Distance between two points:

Suppose that we have two points, $(x_1,y_1)$ and $(x_2,y_2)$. The distance between them is given by:

$D = \sqrt{(x_2-x_1)^2+(y_2-y_1)^2}$

Equation of a circle:

The equation of a circle with center $(x_0,y_0)$ and radius r has the following format:

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

Center at the origin;

This means that $x_0 = 0, y_0 = 0$

So

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

$(x - 0)^2 + (y - 0)^2 = r^2$

$x^2 + y^2 = r^2$

Passes through (4, 6)

The radius is the distance from this point to the center. So

$r = \sqrt{(x_2-x_1)^2+(y_2-y_1)^2}$

$r = \sqrt{(4-0)^2+(6-0)^2}$

$r = \sqrt{16+36}$

$r = \sqrt{52}$

So

$r^2 = 52$

Then

$x^2 + y^2 = r^2$

$x^2 + y^2 = 52$