Question

find the standard form of the equation for the circle with the following properties. endpoints of a diameter are (-5,9) and (-1,-11)

Explanation:

Step1: Find the center of the circle

The center of the circle is the mid - point of the diameter. The mid - point formula for two points $(x_1,y_1)$ and $(x_2,y_2)$ is $(\frac{x_1 + x_2}{2},\frac{y_1 + y_2}{2})$. Here, $x_1=-5,y_1 = 9,x_2=-1,y_2=-11$.
$h=\frac{-5+( - 1)}{2}=\frac{-5 - 1}{2}=\frac{-6}{2}=-3$
$k=\frac{9+( - 11)}{2}=\frac{9 - 11}{2}=\frac{-2}{2}=-1$
So, the center of the circle $(h,k)=(-3,-1)$.

Step2: Find the radius of the circle

The radius $r$ is the distance between the center $(h,k)$ and one of the endpoints of the diameter. We use the distance formula $d=\sqrt{(x_2 - x_1)^2+(y_2 - y_1)^2}$. Let's use the center $(-3,-1)$ and the point $(-5,9)$.
$r=\sqrt{(-5-( - 3))^2+(9-( - 1))^2}=\sqrt{(-5 + 3)^2+(9 + 1)^2}=\sqrt{(-2)^2+10^2}=\sqrt{4 + 100}=\sqrt{104}=2\sqrt{26}$

Step3: Write the standard form of the equation of the circle

The standard form of the equation of a circle is $(x - h)^2+(y - k)^2=r^2$. Substituting $h=-3,k=-1,r = 2\sqrt{26}$ into the formula:
$(x+3)^2+(y + 1)^2=(2\sqrt{26})^2$
$(x + 3)^2+(y + 1)^2=104$

Answer:

$(x + 3)^2+(y + 1)^2=104$