Chapter 9.2, Problem 40WE

To determine

To find: The equation of the circle.

Answer to Problem 40WE

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

Explanation of Solution

Given information: Center on line $x+y=4$ ; and tangent to both coordinate axes.

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

Calculation: Since the circle is tangent to both axes. So, the point can be $\left(x,0\right)$ and $\left(0,y\right)$ .

Now, substitute these coordinates in $x+y=4$ to get center $\left(h,k\right)$ ,

  $\begin{array}{l}x+\left(x+0\right)=4:\left[\because x=\frac{4}{2}=2\right]\\ 2+y=4:y=2\\ (h,k)=\left(2,2\right)\end{array}$

Now, apply the distance formula to get the value of radius,

  $\begin{array}{l}r=\sqrt{\left(x_2-x_1\right)+{\left(y_2-y_1\right)}^2}\\ r=\sqrt{{\left(2-0\right)}^2+{\left(2-2\right)}^2}\\ r=\sqrt{4+0}\\ r=2\end{array}$

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.

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

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

The graph is given below,