Question

Find an equation of the circle that satisfies the given conditions.

The circle lies in the first quadrant tangent to both x-and y-axes; radius5

Short answer

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

Step-by-step solution

Step 1. Determining general Equation of circle.

An equation of the circle with center$(h,k)$ and radiusis

${(x-h)}^2+{(y-k)}^2=r^2$

This is called the standard form for the equation of the circle.

Step 2. Determining the center of the circle.

Since the circle is a tangent to the x-axis

So, radius equals the distance from the y-coordinate of the center to$y=0$

$$\begin{gathered} x-0=5 \\ or,x=5 \end{gathered}$$

Because the circle is a tangent to the x-axis

So, radius equals the distance from the x-coordinate of the center to$x=0$

$\begin{array}{c}y-0=5\\ or,y=5\end{array}$

So, the center of the circle is$(5,5)$

Step 3. Determining the equation of a circle.

Here,$h=5,k=5,r=5$

The equation of the circle is

${(x-5)}^2+{(y-5)}^2=25$