Chapter 1.8, Problem 99E

To determine

To calculate: The equation of a circle whose center is at $\left(7,-3\right)$ and is tangent to x -axis.

Answer to Problem 99E

The equation of the circle is ${\left(x-7\right)}^2+{\left(y+3\right)}^2=9$ .

Explanation of Solution

Given information:

The center of the circle is at $\left(7,-3\right)$ and is tangent to x -axis.

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 $d\left(A,B\right)=\sqrt{{\left(x_2-x_1\right)}^2+{\left(y_2-y_1\right)}^2}$ .

Calculation:

Consider the provided conditions that center of the circle is at $\left(7,-3\right)$ and is tangent to x -axis.

Since the circle is tangent to x -axis so it touches the x -axis at a single point and x -coordinate of the center is same as the x -coordinates of point on x -axis.

Also y coordinate will be zero when circle is tangent to x -axis.

So, the circle passes through the point $\left(7,0\right)$ .

Now, the distance between the point through which the circle passes and the center of the circle is radius.

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 $\left(7,0\right)$ and $\left(7,-3\right)$ .

  $\begin{array}{c}d=\sqrt{{\left(7-7\right)}^2+{\left(-3-0\right)}^2}\\ =\sqrt{9}\\ =3\end{array}$

Therefore, radius of circle is 3.

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 $\left(7,-3\right)$ also radius r is 3.

Here, $h=7,k=-3\text{ and }r=3$ .

Substitute the values in standard equation of circle,

  $\begin{array}{c}{\left(x-7\right)}^2+{\left(y-\left(-3\right)\right)}^2=3^2\\ {\left(x-7\right)}^2+{\left(y+3\right)}^2=9\end{array}$

Thus, the equation of circle is ${\left(x-7\right)}^2+{\left(y+3\right)}^2=9$ .