Question

the points (-3, -1) and (9, 4) are the endpoints of the diameter of a circle. find the equation of the circle.

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 $(x_1,y_1)=(-3,-1)$ and $(x_2,y_2)=(9,4)$.
The x - coordinate of the center $h=\frac{-3 + 9}{2}=\frac{6}{2}=3$.
The y - coordinate of the center $k=\frac{-1 + 4}{2}=\frac{3}{2}$. So the center of the circle is $(3,\frac{3}{2})$.

Step2: Find the radius of the circle

The radius $r$ is the distance between the center $(h,k)=(3,\frac{3}{2})$ and one of the endpoints, say $(x_1,y_1)=(-3,-1)$. The distance formula is $d=\sqrt{(x_2 - x_1)^2+(y_2 - y_1)^2}$.
$r=\sqrt{(3+3)^2+(\frac{3}{2}+1)^2}=\sqrt{6^2+(\frac{3 + 2}{2})^2}=\sqrt{36+\frac{25}{4}}=\sqrt{\frac{144 + 25}{4}}=\sqrt{\frac{169}{4}}=\frac{13}{2}$.

Step3: Write the equation of the circle

The standard form of the equation of a circle is $(x - h)^2+(y - k)^2=r^2$. Substituting $h = 3$, $k=\frac{3}{2}$ and $r=\frac{13}{2}$ we get:
$(x - 3)^2+(y-\frac{3}{2})^2=\frac{169}{4}$.

Answer:

$(x - 3)^2+(y-\frac{3}{2})^2=\frac{169}{4}$