Respuesta :
Answer::
[tex](x-2)^2+y^2=212[/tex]Explanation:
Given the endpoints of the diameter of a circle, we are required to write the equation of the circle.
The standard form of the equation of a circle is given as:
[tex](x-a)^2+(y-b)^2=r^2\text{ where }\begin{cases}Center=(a,b) \\ Radius=r\end{cases}[/tex]So, we need to find two things, the center, and the radius.
Center
The center of the circle is the midpoint of the diameter.
Given the endpoints as: (16, -4) and (-12,4)
[tex]\text{Midpoint}=(\frac{16+(-12)}{2},\frac{-4+4}{2})=(\frac{4}{2},\frac{0}{2})=(2,0)[/tex]The center, (a,b)=(2,0).
Radius
The radius is the distance between the center, (2,0) and one of the endpoints of the diameter.
Using the distance formula, we find the distance between (2,0) and (16,-4).
[tex]\begin{gathered} Distance=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2} \\ =\sqrt[]{(16-2)^2+(-4-0)^2}=\sqrt[]{(14)^2+(-4)^2}=\sqrt[]{212} \\ \implies\text{Radius}=\sqrt{212} \end{gathered}[/tex]Substitute these values into the equation given earlier:
[tex]\begin{gathered} (x-2)^2+(y-0)^2=\sqrt[]{212}^2\text{ where }\begin{cases}Center,(a,b)=(2,0) \\ Radius,r=\sqrt[]{212}\end{cases} \\ (x-2)^2+y^2=212 \end{gathered}[/tex]The equation of the circle is:
[tex](x-2)^2+y^2=212[/tex]
