Question

How do you write an equation for a circle whose diameter has endpoints $(-2,3)$ and $(4,-1)$?

Answer 1

Step-1: Apply Mid point formulae:

Given, that the points are $(-2,3)$ and $(4,-1)$

It is known that the equation for a circle's standard form is ${(x-a)}^2+{(y-b)}^2=r^2$.

Where, $r$ is the radius and$(a,b)$ are the coordinates for the center.

Considering that the diameter's endpoints have been provided.

The radius will then be the distance from the center to either of the two endpoints and the center will be at the midpoint.

The midpoint can be calculated as:

Midpoint formula

$\left(\frac{1}{2}\left(x_1+x_2\right),\frac{1}{2}\left(y_1+y_2\right)\right)$

where, $\left(x_1,y_1\right)$ and $\left(x_2,y_2\right)$ are two points.

Substitute the values in the formula:

$\begin{array}{rcl}\left(\frac{1}{2}\left(x_1+x_2\right),\frac{1}{2}\left(y_1+y_2\right)\right)& =& \left(\frac{-2+4}{2},\frac{3-1}{2}\right)\\ & =& \left(\frac{2}{2},\frac{2}{2}\right)\\ & =& \left(1,1\right)\end{array}$

Step-2: Find the equation of a circle:

The two points are the center $(1,1)$ and the endpoint $(-2,3)$.

$\begin{array}{rcl}r& =& \sqrt{{\left(-2-1\right)}^2+{\left(3-1\right)}^2}\\ & =& \sqrt{9+4}\\ & =& \sqrt{13}\end{array}$

Now, the circle's equation can be expressed in writing:

$\begin{array}{rcl}{(x-1)}^2+{(y-1)}^2& =& {\left(\sqrt{13}\right)}^2\\ & \Rightarrow & {(x-1)}^2+{(y-1)}^2=13\end{array}$

Hence, the equation of a circle is ${\left(x-1\right)}^2+{(y-1)}^2=13$.