Chapter 1.8, Problem 97E

To determine

To calculate: The equation of a circle whose end points of diameter are $P\left(-1,1\right)\text{ and }Q\left(5,9\right)$ .

Answer to Problem 97E

The equation of the circle is ${\left(x-2\right)}^2+{\left(y-5\right)}^2=25$ .

Explanation of Solution

Given information:

The end points of diameter of the circle are $P\left(-1,1\right)\text{ and }Q\left(5,9\right)$ .

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 $P\left(-1,1\right)\text{ and }Q\left(5,9\right)$ .

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 $P\left(-1,1\right)\text{ and }Q\left(5,9\right)$ .

  $\begin{array}{c}\left(\frac{x_1+x_2}{2},\frac{y_1+y_2}{2}\right)=\left(\frac{-1+5}{2},\frac{1+9}{2}\right)\\ =\left(\frac{4}{2},\frac{10}{2}\right)\\ =\left(2,5\right)\end{array}$

Therefore, coordinates of center are $\left(2,5\right)$ .

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 $\left(2,5\right)$ and $\left(-1,1\right)$ .

  $\begin{array}{c}d=\sqrt{{\left(-1-2\right)}^2+{\left(1-5\right)}^2}\\ =\sqrt{9+16}\\ =\sqrt{25}\\ =5\end{array}$

Therefore, radius of circle is 5.

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 $\left(2,5\right)$ also radius r is 5.

Here, $h=2,k=5\text{ and }r=5$ .

Substitute the values in standard equation of circle,

  $\begin{array}{c}{\left(x-2\right)}^2+{\left(y-5\right)}^2=5^2\\ {\left(x-2\right)}^2+{\left(y-5\right)}^2=25\end{array}$

Thus, the equation of circle is ${\left(x-2\right)}^2+{\left(y-5\right)}^2=25$ .