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Q. 2

Expert-verifiedFound in: Page 556

Book edition
6th

Author(s)
Sullivan

Pages
1200 pages

ISBN
9780321795465

Find an equation for the circle with center at the point $(-5,1)$ and radius 3. Graph this circle.

- The equation of the circle is $\begin{array}{rcl}{(x+5)}^{2}+{(y-1)}^{2}& =& 9\end{array}$.
- The graph is

- The center of the circle is $(-5,1)$.
- The radius is 3.

- The general equation of a circle is ${(x-h)}^{2}+{(x-k)}^{2}={r}^{2}$.
- Substitute $h=-5,k=1$ into the general equation and simplify.

localid="1647407263863" $\begin{array}{rcl}(x-{(-5))}^{2}+{(y-1)}^{2}& =& {3}^{2}\\ {(x+5)}^{2}+{(y-1)}^{2}& =& 9\end{array}$

- Substitute 0 for
*x*and find the y*-*intercepts.

$\begin{array}{rcl}{(0+5)}^{2}+{(y-1)}^{2}& =& 9\\ {(y-1)}^{2}& =& -16\end{array}$

- As the square is a negative value, there are no
*x-*intercepts. - Substitute 0 for y
*-*intercepts.

$\begin{array}{rcl}{(x+5)}^{2}+{(0-1)}^{2}& =& 9{(x+5)}^{2}=8x+5=\pm 2\sqrt{2}\\ x& =& -2\sqrt{2}-5,-2\sqrt{2}+5\\ & & \end{array}$

Use the center and the intercepts to plot the circle.

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