Question

Find the function for the bottom half of the circle $x^2+y^2=9$.

Short answer

The function for the bottom half of the circle is $f\left(x\right)=-\sqrt{9-x^2}$.

Step-by-step solution

Step 1. Write down the given circle.

The given circle is $x^2+y^2=9$.

Step 2. Solve for (y) in terms of (x) from the given circle.

Solve for $\left(y\right)$in terms of $\left(x\right)$,

localid="1644994936835" $$\begin{gathered} x^2+y^2=9....\left(\text{Given}\right) \\ \text{or,}{y}^2=9-x^2 \\ y=\pm \sqrt{9-x^2}\mathrm{....}\left(1\right) \end{gathered}$$

Step 3. DESCRIPTION OF STEPS.

By simplifying the given circle, we obtain $y=\pm \sqrt{9-x^2}$, but for bottom half of the circle$\left(y\right)$ is always negative. Therefore from (1),$y=-\sqrt{9-x^2}$ whose function can be written as $f\left(x\right)=-\sqrt{9-x^2}$.