Question

What is the equation of a circle with diameter endpoints of (-14, 1) and (-10, 9)

Answer 1

Given data:

The first end point of the diameter is (-14, 1).

The second end point of the diameter is (-10, 9)​.

The centre of the circel is,

`\begin{gathered} x=(-14-10)/(2) \\ =(-24)/(2) \\ =-12 \\ y=(1+9)/(2) \\ =5 \end{gathered}`

The diameter of the circle is,

`\begin{gathered} d=\sqrt[]{(-10+14)^2+(9-1)^2} \\ =\sqrt[]{16+6}4 \\ =4\sqrt[]{5} \end{gathered}`

The radius is,

`\begin{gathered} r=\frac{4\sqrt[]{5}}{2} \\ =2\sqrt[]{5} \end{gathered}`

The equation for the circle is,

`\begin{gathered} (x-(-12))^2+(y-5)^2=(2\sqrt[]{5})^2 \\ (x+12)^2+(y-5)^2=20 \end{gathered}`

Thus, the equation for the circle is (x+12)^2 +(y-5)^2 =20.