Chapter 1, Problem 27RE

To determine

The standard form of the equation of the circle for the give endpoints of the diameter.

Answer to Problem 27RE

The standard form of equation is ${\left(x-2\right)}^2+{\left(y+3\right)}^2=13$ .

Explanation of Solution

Given:

The endpoints of the diameters are given as:

(0,0) and (4,-6)

Since, the diameter is the double of the length of the radius of the circle and the midpoint of the radius is equal to the centre of the circle.

Evaluating the length of the diameter using the distance formula:

  $\begin{array}{c}d=\sqrt{{\left(4-0\right)}^2+{\left(-6-0\right)}^2}\\ =\sqrt{16+36}\\ =\sqrt{52}\\ =2\sqrt{13}\\ \text{Hence, radius:}\\ r=\frac{d}{2}\\ =\frac{2\sqrt{13}}{2}\\ =\sqrt{13}\end{array}$

Consider the coordinate of the circle are ( a,b ) and it is expressed as:

  $\begin{array}{c}\left(a,b\right)=\left(\frac{x_1+x_2}{2},\frac{y_1+y_2}{2}\right)\\ =\left(\frac{0+4}{2},\frac{0-6}{2}\right)\\ =\left(2,-3\right)\end{array}$

The general equation of the circle is given for the coordinate ( a,b ) is given as:

  $\begin{array}{l}{\left(x-a\right)}^2+{\left(y-b\right)}^2=r^2\\ \text{Here,}\\ {\left(x-2\right)}^2+{\left(y+3\right)}^2={\left(\sqrt{13}\right)}^2\\ {\left(x-2\right)}^2+{\left(y+3\right)}^2=13\end{array}$