Question

determining the equation given the center and a point on the circle
determine the equation of a circle with a center at (-4, 0) that passes through the point (-2, 1) by following the steps below.

1. use the distance formula to determine the radius:

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

1. substitute the known values into the standard form:

$(x - h)^2 + (y - k)^2 = r^2$.
what is the equation of a circle with a center at (-4, 0) that passes through the point (-2, 1)?
$\bigcirc$ $x^2 + (y + 4)^2 = \sqrt{5}$
$\bigcirc$ $(x - 1)^2 + (y + 2)^2 = 5$
$\bigcirc$ $(x + 4)^2 + y^2 = 5$
$\bigcirc$ $(x + 2)^2 + (y - 1)^2 = \sqrt{5}$

Explanation:

Step1: Calculate the radius

The center is $(h,k)=(-4,0)$ and the point on the circle is $(x_2,y_2)=(-2,1)$. Use the distance formula:

$$ r = \sqrt{(-2 - (-4))^2 + (1 - 0)^2} = \sqrt{(2)^2 + (1)^2} = \sqrt{4 + 1} = \sqrt{5} $$

Then square the radius for the standard form: $r^2 = (\sqrt{5})^2 = 5$

Step2: Substitute into standard form

The standard circle equation is $(x-h)^2+(y-k)^2=r^2$. Substitute $h=-4$, $k=0$, $r^2=5$:

$$ (x - (-4))^2 + (y - 0)^2 = 5 \implies (x+4)^2 + y^2 = 5 $$

Answer:

$\boldsymbol{(x + 4)^2 + y^2 = 5}$ (corresponding to the third option)