Question

How do you write an equation for a circle with endpoints of a diameter at the points (7,4) and (-9,6)?

Answer 1

The endpoints of a diameter of the circle are given as (7, 4) and (-9, 6).

To write the equation of the circle, we have to get the coordinates of the center of the circle.

We know the diameter passes through the center of the circle, and also the center of the circle is the midpoint of every diameter of the circle.

So we have to use the concept below to get the coordinates of the center of the circle. ↓↓↓

$\boxed{\begin{minipage}{11 cm}The midpoint of line segment joining points (x_1,\ y_1) and (x_2,\ y_2) \\ \\ =\left(\dfrac{x_1+x_2}{2},\ \dfrac{y_1+y_2}{2}\right) \end{minipage}}$

Thus the center of the circle is,

$\begin{aligned}&\left(\frac{7+(-9)}{2},\ \frac{4+6}{2}\right)\\ \\ \Longrightarrow\ \ &\left(\frac{7-9}{2},\ \frac{10}{2}\right)\\ \\ \Longrightarrow\ \ &\left(\frac{-2}{2},\ 5\right)\\ \\ \Longrightarrow\ \ &\left(-1,\ 5\right)\end{aligned}$

On considering the center of the circle (-1, 5) and the point (7, 4) which is on the circle, the radius of the circle is,

$\begin{aligned}&\sqrt{(-1-7)^2+(5-4)^2}\\ \\ \Longrightarrow\ \ &\sqrt{(-8)^2+1^2}\\ \\ \Longrightarrow\ \ &\sqrt{64+1}\\ \\ \Longrightarrow\ \ &\sqrt{65}\end{aligned}$

Let  (x, y)  be a point on the circle.

On considering this (x, y) and the center of the circle (-1, 5), the radius of the circle can be written as,

$\begin{aligned}&\sqrt{(-1-x)^2+(5-y)^2}&=&\ \ \sqrt{65}\\ \\ \Longrightarrow\ \ &(-1-x)^2+(5-y)^2&=&\ \ 65\\ \\ \Longrightarrow\ \ &1+2x+x^2+25-10y+y^2&=&\ \ 65\\ \\ \Longrightarrow\ \ &x^2+y^2+2x-10y+26&=&\ \ 65\\ \\ \Longrightarrow\ \ &x^2+y^2+2x-10y+26-65&=&\ \ 0\\ \\ \Longrightarrow\ \ &\Large x^2+y^2+2x-10y-39&\Large =&\ \ \Large 0\end{aligned}$

This is the equation of the circle!