Question

Identify the equation of the circle that has its center at (-3, -4) and passes through the origin.

Answer 1

The equation of a circle:

$(x-h)^2+(y-k)^2=r^2$

(h, k) - a center

r - a radius

We have:

The center (-3, -4) → h = -3 and k = -4.

The radius is equal the distance between the center of the circle and the origin.

The formula of a distance between two points:

$A(x_A,\ y_A),\ B(X_B,\ y_B)\\\\d_{AB}=\sqrt{(x_B-x_A)^2+(y_B-y_A)^2}$

Substitute

$(-3,\ -4),\ (0,\ 0)\\\\d=\sqrt{(0-(-3))^2+(0-(-4))^2}=\sqrt{9+16}=\sqrt{25}=5$

therefore r = 5.

Answer:

$(x-(-3))^2+(y-(-4))^2=5^2\\\\(x+3)^2+(y+4)^2=25$