Question

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. $d = \sqrt{(x_2 - x_1)^2+(y_2 - y_1)^2}$.
2 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)?
$x^2+(y + 4)^2=\sqrt{5}$
$(x - 1)^2+(y + 2)^2 = 5$
$(x + 4)^2+y^2 = 5$
$(x + 2)^2+(y - 1)^2=\sqrt{5}$

Explanation:

Step1: Calculate the radius

Let $(x_1,y_1)=(-4,0)$ and $(x_2,y_2)=(-2,1)$. Using the distance formula $r = \sqrt{(x_2 - x_1)^2+(y_2 - y_1)^2}$, we have $r=\sqrt{(-2-(-4))^2+(1 - 0)^2}=\sqrt{(2)^2+(1)^2}=\sqrt{4 + 1}=\sqrt{5}$.

Step2: Find the equation of the circle

The standard - form of a circle is $(x - h)^2+(y - k)^2=r^2$, where $(h,k)$ is the center of the circle. Here, $h=-4,k = 0,r=\sqrt{5}$. Substituting these values, we get $(x-(-4))^2+(y - 0)^2=(\sqrt{5})^2$, which simplifies to $(x + 4)^2+y^2=5$.

Answer:

$(x + 4)^2+y^2=5$