Question

Find the equation of the circle that has a diameter with endpoints located at -3,6) and 9,6). - A. (x-6)² + (y - 3)² = 36 B. (x+3)² + (y+6)²= 12 C. (x – 3)²+(y-6)² = 36 D. (x – 3)²+ (y-6)² = 144 -

Answer 1

Take into account that the general equation of a circumference is given by:

`(x-h)^2+(y-k)^2=r^2`

where (h,k) is the center of the circle and r the radius.

To determine the center of the circle calculate the coordinates of the midpoint between the endpoints of the diameters:

`\begin{gathered} x=(x_1+x_2)/(2)=(-3+9)/(2)=(6)/(2)=3 \\ y=(y_1+y_2)/(2)=(6+6)/(2)=(12)/(2)=6 \end{gathered}`

hence, the midpoint is (h,k) = (3,6) and it is the center of the circle.

To determine the radius of the circle, find first the lengh of the diameter, as follow:

`\begin{gathered} d=\sqrt[]{(-3-9)^2+(6-6)^2} \\ d=\sqrt[]{144}=12 \end{gathered}`

Then, the radius is:

r = d/2 = 12/2 = 6

By replacing the previous values of h, k and r into the general equation for a circle, you obtain:

`\begin{gathered} (x-3)^2+(y-6)^2=6^2 \\ (x-3)^2+(y-6)^2=36 \end{gathered}`