Question

find the standard form of the equation for the circle with the following properties. center (-8,9), passes through (-1,0)

Explanation:

Step1: Recall the standard - form of a circle equation

The standard - form of a circle equation 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 $(h,k)=(-8,9)$, so the equation is $(x + 8)^2+(y - 9)^2=r^2$.

Step2: Calculate the radius

The radius $r$ is the distance between the center $(-8,9)$ and the point $(-1,0)$ on the circle. Use the distance formula $d=\sqrt{(x_2 - x_1)^2+(y_2 - y_1)^2}$. Let $(x_1,y_1)=(-8,9)$ and $(x_2,y_2)=(-1,0)$. Then $r=\sqrt{(-1+8)^2+(0 - 9)^2}=\sqrt{7^2+(-9)^2}=\sqrt{49 + 81}=\sqrt{130}$.

Step3: Write the final equation

Substitute $r^2 = 130$ into the equation $(x + 8)^2+(y - 9)^2=r^2$. The standard - form of the circle equation is $(x + 8)^2+(y - 9)^2=130$.

Answer:

$(x + 8)^2+(y - 9)^2=130$