Question

Find the equation of the circle that has a diameter with endpoints located at (7,3) and (7,-5). A. (x-7)²+(x+ 1)² = 16 B. (x-7)² + (-1)² =4 C. (x+1)²+(y-7)= 16 D. (x-7)²+(y+ 1)²= 64

Answer 1

Answer:

The equation of the circle is;

$\mleft(x-7\mright)^2+\mleft(y+1\mright)^2=16$

Explanation:

Given that the circle has a diameter with endpoints located at (7,3) and (7,-5).

The diameter of the circle is the distance between the two points;

$\begin{gathered} d=\sqrt[]{(7-7)^2+(3--5)^2_{}} \\ d=\sqrt[]{(0)^2+(3+5)^2_{}} \\ d=\sqrt[]{64} \\ d=8 \end{gathered}$

The radius of the circle is;

$\begin{gathered} r=\frac{d}{2}=\frac{8}{2} \\ r=4 \end{gathered}$

The center of the circle is at the midpoint of the line of the diameter.

$\begin{gathered} (h,k)=(\frac{7+7}{2},\frac{3-5}{2}) \\ (h,k)=(\frac{14}{2},\frac{-2}{2}) \\ (h,k)=(7,-1) \end{gathered}$

Applying the equation of a circle;

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

Substituting the given values;

$\begin{gathered} (x-h)^2+(y-k)^2=r^2 \\ (x-7)^2+(y+1)^2=4^2 \\ (x-7)^2+(y+1)^2=16^{} \end{gathered}$

Therefore, the equation of the circle is;

$\mleft(x-7\mright)^2+\mleft(y+1\mright)^2=16$