Question

find the standard form of the equation of the circle with endpoints of a diameter at the points (1,6) and (-7,4)
type the standard form of the equation of this circle.
(type an equation.)

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 $(\frac{x_1 + x_2}{2},\frac{y_1 + y_2}{2})$. Given $(x_1,y_1)=(1,6)$ and $(x_2,y_2)=(-7,4)$.
$h=\frac{1+( - 7)}{2}=\frac{1 - 7}{2}=\frac{-6}{2}=-3$
$k=\frac{6 + 4}{2}=\frac{10}{2}=5$
So the center of the circle is $(h,k)=(-3,5)$.

Step2: Find the radius of the circle

The radius $r$ is the distance from the center $(h,k)$ to either 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 $(-3,5)$ and the point $(1,6)$.
$r=\sqrt{(1-( - 3))^2+(6 - 5)^2}=\sqrt{(1 + 3)^2+(6 - 5)^2}=\sqrt{4^2+1^2}=\sqrt{16 + 1}=\sqrt{17}$

Step3: Write the standard form of the circle equation

The standard form of the equation of a circle is $(x - h)^2+(y - k)^2=r^2$. Substituting $h=-3$, $k = 5$ and $r=\sqrt{17}$ into the equation.
$(x+3)^2+(y - 5)^2=17$

Answer:

$(x + 3)^2+(y - 5)^2=17$