Question

Write the standard equation of the circle with center (-12, – 7) that passes through the point (-3,7).

Answer 1

Answer:

$(x + 12)^2 + (y + 7)^2 = 277$

Step-by-step explanation:

Equation of a circle:

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

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

In which r is the radius.

Center (-12, – 7)

This means that $x_0 = -12, y_0 = -7$. So

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

$(x - (-12))^2 + (y - (-7))^2 = r^2$

$(x + 12)^2 + (y + 7)^2 = r^2$

Passes through the point (-3,7).

This means that we use $x = -3, y = 7$ to find the radius squared. So

$(x + 12)^2 + (y + 7)^2 = r^2$

$(-3 + 12)^2 + (7 + 7)^2 = r^2$

$81 + 196 = r^2$

$r^2 = 277$

The equation of the circle is:

$(x + 12)^2 + (y + 7)^2 = r^2$

$(x + 12)^2 + (y + 7)^2 = 277$