Chapter 1, Problem 106RE

To determine

To calculate:The equation for the given circle which has points $P\left(2,3\right)$ and $Q\left(-1,8\right)$ . Also, its center is the midpoint of the given segment $PQ$

Answer to Problem 106RE

The equation for the circle thathas points $P\left(2,3\right)$ and $Q\left(-1,8\right)$ , center is the midpoint of the given segment $PQ$ is $x^2+y^2-x-11y+22=0$

Explanation of Solution

Given information:

A circle withpoints $P\left(2,3\right)$ and $Q\left(-1,8\right)$ lying inside it and whose center is given by midpoint of $PQ$

Formula used:

For a given circle with center $\left(h,k\right)$ and radius $r$ the equation is given as:

  ${\left(x-h\right)}^2+{\left(y-k\right)}^2=r^2$

This is referred to as the Standard form for the equation of a given circle.

The Distance denoted by say, $D$ between any two given points say $\left(x_1,y_1\right)$ and $\left(x_2,y_2\right)$ is given as:

  $D=\sqrt{{\left(x_2-x_1\right)}^2+{\left(y_2-y_1\right)}^2}$

The midpoint $\left(x,y\right)$ for two points $\left(x_1,y_1\right)$ and $\left(x_2,y_2\right)$ is given as:

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

The relation between $d$ , diameter of circle and $r$ , radius of circle is:

  $r=\frac{d}{2}$

Calculation:

Consider the given circle with points $P\left(2,3\right)$ and $Q\left(-1,8\right)$ lying inside it.

Then in order to apply the midpoint formula for the segment $PQ$ the following should be known:

  $\begin{array}{l}{x}_1=2\\ x_2=-1\\ y_1=3\\ y_2=8\end{array}$

Recall the midpoint $\left(x,y\right)$ for two points $\left(x_1,y_1\right)$ and $\left(x_2,y_2\right)$ is given as:

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

Therefore,

  $\begin{array}{l}\left(x,y\right)=\left(\frac{2+\left(-1\right)}{2},\frac{3+8}{2}\right)\\ =\left(\frac{1}{2},\frac{11}{2}\right)\end{array}$

Hence, center of the circle is $\left(x,y\right)=\left(\frac{1}{2},\frac{11}{2}\right)$

Now as midpoint of $PQ$ forms the center of the circle, therefore length of $PQ$ gives the diameter for the circle.

And the relation between $d$ , diameter of circle and $r$ , radius of circle is:

  $r=\frac{d}{2}$

Therefore, distance from center $\left(x,y\right)=\left(\frac{1}{2},\frac{11}{2}\right)$ to $P\left(2,3\right)$ gives the radius for the circle.

Recall the distance $D$ between two points $\left(x_1,y_1\right)$ and $\left(x_2,y_2\right)$ is given as:

  $D=\sqrt{{\left(x_2-x_1\right)}^2+{\left(y_2-y_1\right)}^2}$

As we have:

  $\begin{array}{l}{x}_1=\frac{1}{2}\\ x_2=2\\ y_1=\frac{11}{2}\\ y_2=3\end{array}$

Therefore,

  $\begin{array}{l}D=\sqrt{{\left(2-\frac{1}{2}\right)}^2+{\left(3-\frac{11}{2}\right)}^2}\\ =\sqrt{{\left(\frac{3}{2}\right)}^2+{\left(\frac{-5}{2}\right)}^2}\\ =\sqrt{\frac{9}{4}+\frac{25}{4}}\\ =\sqrt{\frac{34}{4}}\\ =\frac{\sqrt{34}}{2}\end{array}$

Hence, radius $r=\frac{\sqrt{34}}{2}$

Also we have center $\left(h,k\right)=\left(\frac{1}{2},\frac{11}{2}\right)$

Recall that for a given circle with center $\left(h,k\right)$ and radius $r$ the equation is given as:

  ${\left(x-h\right)}^2+{\left(y-k\right)}^2=r^2$

Therefore, replacing the values we get:

  ${\left(x-\frac{1}{2}\right)}^2+{\left(y-\frac{11}{2}\right)}^2={\left(\frac{\sqrt{34}}{2}\right)}^2$ $\left(1\right)$

Now using the following formulae:

  $\begin{array}{l}{\left(a+b\right)}^2=a^2+2ab+b^2\\ {\left(a-b\right)}^2=a^2-2ab+b^2\end{array}$

Then, $\left(1\right)$ reduces to:

  $\begin{array}{l}{\left(x-\frac{1}{2}\right)}^2+{\left(y-\frac{11}{2}\right)}^2={\left(\frac{\sqrt{34}}{2}\right)}^2\\ x^2+\frac{1}{4}-x+y^2+\frac{121}{4}-11y=\frac{34}{4}\\ x^2+y^2-x-11y+\left(\frac{1}{4}+\frac{121}{4}-\frac{34}{4}\right)=0\\ x^2+y^2-x-11y+\left(\frac{88}{4}\right)=0\\ x^2+y^2-x-11y+22=0\end{array}$

Hence, the equation for the circle that has points $P\left(2,3\right)$ and $Q\left(-1,8\right)$ , center is the midpoint of the given segment $PQ$ is $x^2+y^2-x-11y+22=0$