Question

write the standard form of the equation of the circle having the given center and containing the given point. center: (-2,0); point: (3,2)

Explanation:

Step1: Recall circle - equation formula

The standard form of the equation of a circle is $(x - h)^2+(y - k)^2=r^2$, where $(h,k)$ is the center of the circle and $r$ is the radius. Here, the center $(h,k)=(-2,0)$.

Step2: Calculate the radius

The radius $r$ is the distance between the center $(-2,0)$ and the point $(3,2)$ on the circle. Use the distance formula $d=\sqrt{(x_2 - x_1)^2+(y_2 - y_1)^2}$. Let $(x_1,y_1)=(-2,0)$ and $(x_2,y_2)=(3,2)$. Then $r=\sqrt{(3+2)^2+(2 - 0)^2}=\sqrt{25 + 4}=\sqrt{29}$.

Step3: Write the equation of the circle

Substitute $h=-2$, $k = 0$, and $r=\sqrt{29}$ into the standard - form equation. We get $(x+2)^2+(y - 0)^2=(\sqrt{29})^2$.

Answer:

$(x + 2)^2+y^2=29$