Question

find the standard form of the equation of the circle with endpoints of a diameter at the points (9,6) and (-7,8). 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 $(\frac{x_1 + x_2}{2},\frac{y_1 + y_2}{2})$. Here, $x_1=9,y_1 = 6,x_2=-7,y_2 = 8$. So the center $(h,k)=(\frac{9+( - 7)}{2},\frac{6 + 8}{2})=(1,7)$.

Step2: Find the radius of the circle

The radius $r$ is the distance between the center $(h,k)=(1,7)$ and one of the endpoints of the diameter, say $(x_1,y_1)=(9,6)$. The distance formula is $r=\sqrt{(x_1 - h)^2+(y_1 - k)^2}$. So $r=\sqrt{(9 - 1)^2+(6 - 7)^2}=\sqrt{64 + 1}=\sqrt{65}$.

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 $h = 1,k = 7,r=\sqrt{65}$, we get $(x - 1)^2+(y - 7)^2=65$.

Answer:

$(x - 1)^2+(y - 7)^2=65$