Chapter 9.2, Problem 41WE

To determine

To find: The equation of the circle.

Answer to Problem 41WE

The required equation is ${\left(x-5\right)}^2+{\left(y+4\right)}^2=16$ .

Explanation of Solution

Given information: Tangent to the line $x=1,x=9$ and $y=0$

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

Calculation: Since tangents $x=1andx=9$ are parallel. This means the endpoints of the diameter of the circle touch these lines. Hence, the coordinates of the endpoints are $\left(1,y\right)and\left(9,y\right)$ . Where $\text{'}y\text{'}$ is an unknown variable.

As we have the end x-coordinate of diameter,

Hence, the midpoint of the diameter .i.e. center of the circle $=\frac{1+9}{2}=5$

  $\begin{array}{l}\because x=5\\ \therefore h=5\left\{\because x=h\right\}\end{array}$

The distance from 1 to 5 is 4 and the distance from 5 to 9 is 4.

Hence, the radius of the circle $=r=4$ .

Since, the given circle is also tangent to $y=0$ . It means the center of the circle is either $-4$ or $4$ units away from $\left(x,0\right)$ .

It is also mention in question that the circle lies in the IV quadrant. Hence, $4$ is negligible.

Now, the center of the circle will be $\left(5,-4\right)$ .

Substitute the value of the center $\left(h,k\right)$ and the radius $\left(r\right)$ into ${\left(x-h\right)}^2+{\left(y-k\right)}^2=r^2$ to get the equation of the circle.

  $\begin{array}{l}{\left(x-h\right)}^2+{\left(y-k\right)}^2=r^2\\ {\left(x-5\right)}^2+{\left(y-\left(-4\right)\right)}^2=4^2\\ {\left(x-5\right)}^2+{\left(y+4\right)}^2=16\end{array}$

Therefore, the required equation is ${\left(x-5\right)}^2+{\left(y+4\right)}^2=16$ .

The graph is given below,