Question

Write the standard equation of the circle with center (-14, -3) that passes through the point (-5,3),

Answer 1

Answer:

$(x + 14)^2 + (y + 3)^2 = 117$

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 (-14, -3)

This means that $x_0 = -14, y_0 = -3$

So

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

$(x - (-14))^2 + (y - (-3))^2 = r^2$

$(x + 14)^2 + (y + 3)^2 = r^2$

Passes through the point (-5,3),

This means that when $x = -5, y = 3$. We use this to find the radius squared. So

$(x + 14)^2 + (y + 3)^2 = r^2$

$(-5 + 14)^2 + (3 + 3)^2 = r^2$

$r^2 = 117$

So, the equation of the circle is:

$(x + 14)^2 + (y + 3)^2 = r^2$

$(x + 14)^2 + (y + 3)^2 = 117$