Question

3.) find an equation of the circle that has endpoints of a diameter a(8, 10) and b(-2, -14).

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 $(\frac{x_1 + x_2}{2},\frac{y_1 + y_2}{2})$. For points $A(8,10)$ and $B(-2,-14)$, the center $(h,k)$ is $(\frac{8+( - 2)}{2},\frac{10+( - 14)}{2})=(3,-2)$.

Step2: Find the radius of the circle

The radius $r$ is the distance from the center $(h,k)$ to either endpoint of the diameter. Using the distance formula $d=\sqrt{(x_2 - x_1)^2+(y_2 - y_1)^2}$, with $(h,k)=(3,-2)$ and $(x_1,y_1)=(8,10)$:
\[

$$\begin{align*} r&=\sqrt{(8 - 3)^2+(10-( - 2))^2}\\ &=\sqrt{5^2+12^2}\\ &=\sqrt{25 + 144}\\ &=\sqrt{169}\\ & = 13 \end{align*}$$

\]

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 $h = 3,k=-2,r = 13$ into the equation, we get $(x - 3)^2+(y + 2)^2=169$.

Answer:

$(x - 3)^2+(y + 2)^2=169$