Chapter 1.8, Problem 98E

To determine

To calculate: The equation of a circle whose end points of diameter are $P\left(-1,3\right)\text{ and }Q\left(7,-5\right)$ .

Answer to Problem 98E

The equation of the circle is ${\left(x-3\right)}^2+{\left(y+1\right)}^2=32$ .

Explanation of Solution

Given information:

The end points of diameter of the circle are $P\left(-1,3\right)\text{ and }Q\left(7,-5\right)$ .

Formula used:

The standard form of the equation of the circle is ${\left(x-h\right)}^2+{\left(y-k\right)}^2=r^2$ , where $\left(h,k\right)$ denote the center of the circle and r denote the radius.

Distance $d\left(A,B\right)$ between two points $A=\left(x_1,y_1\right)\text{ and }B=\left(x_2,y_2\right)$ in the Cartesian plane is denoted by $d\left(A,B\right)=\sqrt{{\left(x_2-x_1\right)}^2+{\left(y_2-y_1\right)}^2}$ .

Midpoint of the segment that joins two points $A=\left(x_1,y_1\right)\text{ and }B=\left(x_2,y_2\right)$ in the Cartesian plane is denoted by $\left(\frac{x_1+x_2}{2},\frac{y_1+y_2}{2}\right)$ .

Calculation:

Consider the provided conditions that end points of diameter of the circle are $P\left(-1,3\right)\text{ and }Q\left(7,-5\right)$ .

Now, the center of the circle is the midpoint of the diameter.

Recall that the midpoint of the segment that joins two points $A=\left(x_1,y_1\right)\text{ and }B=\left(x_2,y_2\right)$ in the Cartesian plane is denoted by $\left(\frac{x_1+x_2}{2},\frac{y_1+y_2}{2}\right)$ .

Evaluate the midpoint of the segment that joins $P\left(-1,3\right)\text{ and }Q\left(7,-5\right)$ .

  $\begin{array}{c}\left(\frac{x_1+x_2}{2},\frac{y_1+y_2}{2}\right)=\left(\frac{-1+7}{2},\frac{3-5}{2}\right)\\ =\left(\frac{6}{2},\frac{-2}{2}\right)\\ =\left(3,-1\right)\end{array}$

Therefore, coordinates of center are $\left(3,-1\right)$ .

Now, the radius of the circle is distance between the center of the circle and any one end point of diameter.

Recall that the distance $d\left(A,B\right)$ between two points $A=\left(x_1,y_1\right)\text{ and }B=\left(x_2,y_2\right)$ in the Cartesian plane is denoted by $d\left(A,B\right)=\sqrt{{\left(x_2-x_1\right)}^2+{\left(y_2-y_1\right)}^2}$ .

Evaluate the distance between $\left(3,-1\right)$ and $\left(-1,3\right)$ .

  $\begin{array}{c}d=\sqrt{{\left(-1-3\right)}^2+{\left(3+1\right)}^2}\\ =\sqrt{16+16}\\ =\sqrt{32}\end{array}$

Therefore, radius of circle is $\sqrt{32}$ .

Recall that the standard form of the equation of the circle is ${\left(x-h\right)}^2+{\left(y-k\right)}^2=r^2$ , where $\left(h,k\right)$ denote the center of the circle and r denote the radius.

Compare, $\left(h,k\right)$ and $\left(3,-1\right)$ also radius r is $\sqrt{32}$ .

Here, $h=3,k=-1\text{ and }r=\sqrt{32}$ .

Substitute the values in standard equation of circle,

  $\begin{array}{c}{\left(x-3\right)}^2+{\left(y-\left(-1\right)\right)}^2={\left(\sqrt{32}\right)}^2\\ {\left(x-3\right)}^2+{\left(y+1\right)}^2=32\end{array}$

Thus, the equation of circle is ${\left(x-3\right)}^2+{\left(y+1\right)}^2=32$ .