Question Statement

A circle touches both the $x$-axis and $y$-axis in the first quadrant. Given that the area of the circle is $13\pi$ square units, find the equation of the circle.

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Background and Explanation

To find the equation of a circle, we need to determine its center $(h,k)$ and its radius $r$. When a circle touches both axes in the first quadrant, the distance from the center to each axis is equal to the radius, meaning the center is located at $(r,r)$.

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Solution

Step 1: Determine the radius of the circle

We are given that the area of the circle is $13\pi$ square units. Using the area formula $A=\pi r^2$:

$\begin{array}{rl}\pi r^2& =13\pi \\ r^2& =13\\ r& =\sqrt{13}\end{array}$

Thus, the radius of the circle is $\sqrt{13}$.

Step 2: Determine the center of the circle

Since the circle touches both the $x$-axis and the $y$-axis in the first quadrant, the coordinates of the center are equal to the radius.

Therefore, the center of the circle is $(h,k)=(\sqrt{13},\sqrt{13})$.

Step 3: Find the equation of the circle

The standard form of the equation of a circle is $(x-h{)}^2+(y-k{)}^2=r^2$. Substituting the values we found:

$\begin{array}{rl}(x-\sqrt{13}{)}^2+(y-\sqrt{13}{)}^2& =(\sqrt{13}{)}^2\\ \Rightarrow x^2-2\sqrt{13}x+13+y^2-2\sqrt{13}y+13& =13\\ \Rightarrow x^2+y^2-2\sqrt{13}x-2\sqrt{13}y+13+13-13& =0\\ \Rightarrow x^2+y^2-2\sqrt{13}x-2\sqrt{13}y+13& =0\end{array}$

The final equation of the circle is $x^2+y^2-2\sqrt{13}x-2\sqrt{13}y+13=0$.

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Key Formulas or Methods Used

• Area of a Circle: $A=\pi r^2$
• Circle Touching Both Axes (First Quadrant): Center is $(r,r)$ and radius is $r$.
• Standard Equation of a Circle: $(x-h{)}^2+(y-k{)}^2=r^2$
• Algebraic Expansion: $(a-b{)}^2=a^2-2ab+b^2$

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Summary of Steps

1. Use the provided area ($13\pi$) to calculate the radius $r=\sqrt{13}$.
2. Identify that for a circle touching both axes in the first quadrant, the center is $(r,r)$, which is $(\sqrt{13},\sqrt{13})$.
3. Plug the center and radius into the standard circle equation formula.
4. Expand the binomials and simplify the equation into the general form $x^2+y^2+Dx+Ey+F=0$.