Practice 2 Solutions

4. Write the equation of the circle in the form ${\left(x–h\right)}^2+{\left(y–k\right)}^2=r^2$ with a center of $(10.45, –23.4)$ and a radius of 12.3 units.

${\left(x–10.45\right)}^2+{\left(y–\left(–23.4\right)\right)}^2=12.3^2$

${\left(x–10.45\right)}^2+{\left(y+23.4\right)}^2=151.29$

5. Write the equation of the circle in the form ${\left(x–h\right)}^2+{\left(y–k\right)}^2=r^2$ with a center of $\left(–\sqrt{2}, 4\sqrt{3}\right)$ and a radius of $\sqrt{10}$.

${\left(x–\left(–\sqrt{2}\right)\right)}^2+{\left(y–4\sqrt{3}\right)}^2={\left(\sqrt{10}\right)}^2$

$\left(x+\sqrt{2}\right)+{\left(y–4\sqrt{3}\right)}^2=10$

6. Name the domain and range of the circle given the equation: ${\left(x+2\right)}^2+{\left(y–4\right)}^2=196$

$$\begin{gathered} \mathrm{center}: \left(–2, 4\right) \\ \mathrm{radius}: 14 \\ \mathrm{domain}: –2\pm 14 \\ \mathrm{range}: 4\pm 14 \end{gathered}$$

$$\begin{gathered} \mathrm{domain}: \left\{x|x\in \mathrm{ℝ}, –16\le x\le 12\right\} \\ \mathrm{range}: \left\{y|y\in \mathrm{ℝ}, –10\le y\le 18\right\} \end{gathered}$$

7. Name the domain and range of the circle given the equation: ${\left(x–1.32\right)}^2+{\left(y–4.13\right)}^2=5.76$

$$\begin{gathered} \mathrm{center}: \left(1.32, 4.13\right) \\ \mathrm{radius}: 2.4 \\ \mathrm{domain}: 1.32\pm 2.4 \\ \mathrm{range}: 4.13\pm 2.4 \end{gathered}$$

$$\begin{gathered} \mathrm{domain}: \left\{x|x\in \mathrm{ℝ}, –1.08\le x\le 3.72\right\} \\ \mathrm{range}: \left\{y|y\in \mathrm{ℝ}, 1.73\le y\le 6.53\right\} \end{gathered}$$

8. Determine the equation of the circle when the endpoints of the diameter are $(4, 5)$ and $(–9, 3)$. 

$$\begin{gathered} \mathrm{center}: \left(\frac{4+–9}{2}, \frac{5+3}{2}\right)=\left(–\frac{5}{2}, 4\right) \\ \left(–\frac{5}{2}, 4\right), \left(4, 5\right) \\ r=\sqrt{{\left(–\frac{5}{2}–\frac{8}{2}\right)}^2+{\left(4–5\right)}^2} \\ r=\sqrt{{\left(–\frac{13}{2}\right)}^2+{\left(–1\right)}^2} \\ r=\sqrt{\frac{169}{4}+\frac{4}{4}} \\ r=\sqrt{\frac{173}{4}} \\ {\left(x–\left(–\frac{5}{2}\right)\right)}^2+{\left(y–4\right)}^2={\left(\sqrt{\frac{173}{4}}\right)}^2 \end{gathered}$$

${\left(x+\frac{5}{2}\right)}^2+{\left(y–4\right)}^2=\frac{173}{4}$

9. Determine the equation of the circle with the center at $(–5, 2)$ and an endpoint $(2, 1)$. 

$$\begin{gathered} \left(–5, 2\right), \left(2, 1\right) \\ r=\sqrt{{\left(–5–2\right)}^2+{\left(2–1\right)}^2} \\ r=\sqrt{{\left(–7\right)}^2+{\left(1\right)}^2} \\ r=\sqrt{50} \\ {\left(x–\left(–5\right)\right)}^2+{\left(y–2\right)}^2={\left(\sqrt{50}\right)}^2 \end{gathered}$$

${\left(x+5\right)}^2+{\left(y–2\right)}^2=50$

12. Write the equation of the circle in standard form: $\frac{2}{3}{x}^2+3x+\frac{2}{3}{y}^2–7y=1$

$$\begin{gathered} \frac{2}{3}\left(x^2+\frac{9}{2}x\right)+\frac{2}{3}\left(y^2–\frac{21}{2}y\right)=1 \\ \frac{2}{3}\left(x^2+\frac{9}{2}x+\boxed{{\left(\frac{9}{4}\right)}^2}\right)+\frac{2}{3}\left(y^2–\frac{21}{2}y+\boxed{{\left(–\frac{21}{4}\right)}^2}\right)=1+\frac{2}{3}\boxed{{\left(\frac{9}{4}\right)}^2}+\frac{2}{3}\boxed{{\left(–\frac{21}{4}\right)}^2} \\ \frac{2}{3}{\left(x+\frac{9}{4}\right)}^2+\frac{2}{3}{\left(y–\frac{21}{2}\right)}^2=1+\frac{2}{3}\left(\frac{81}{16}\right)+\frac{2}{3}\left(\frac{441}{16}\right) \\ \frac{2}{3}{\left(x+\frac{9}{4}\right)}^2+\frac{2}{3}{\left(y–\frac{21}{4}\right)}^2=1+\frac{27}{8}+\frac{147}{8} \\ \frac{3}{2}\left(\frac{2}{3}{\left(x+\frac{9}{4}\right)}^2+\frac{2}{3}{\left(y–\frac{21}{4}\right)}^2\right)=\left(\frac{91}{4}\right)\frac{3}{2} \end{gathered}$$

${\left(x+\frac{9}{4}\right)}^2+{\left(y–\frac{21}{4}\right)}^2=\frac{273}{8}$

13. Write the equation of a circle in standard form: $5x^2–20x+5y^2+10x=2$

$$\begin{gathered} 5\left(x^2–4x+\boxed{{\left(–\frac{4}{2}\right)}^2}\right)+5\left(y^2+2x+\boxed{{\left(\frac{2}{2}\right)}^2}\right)=2+5\boxed{{\left(–\frac{4}{2}\right)}^2}+5\boxed{{\left(\frac{2}{2}\right)}^2} \\ 5{\left(x–2\right)}^2+5{\left(y+1\right)}^2=2+5\left(4\right)+5\left(1\right) \\ 5{\left(x–2\right)}^2+5{\left(y+1\right)}^2=2+20+5 \\ 5{\left(x–2\right)}^2+5{\left(y+1\right)}^2=27 \end{gathered}$$

${\left(x–2\right)}^2+{\left(y+1\right)}^2=\frac{27}{5}$

14. A food delivery company centered at $(7, 5)$ will deliver within a radius of 6 units. Determine if a house located at $(2, 6)$ is within that radius.

$$\begin{gathered} \left(7, 5\right), \left(2, 6\right) \\ r=\sqrt{{\left(7–2\right)}^2+{\left(5–6\right)}^2} \\ r=\sqrt{{\left(5\right)}^2+{\left(–1\right)}^2} \\ r=\sqrt{26}\approx 5.10 \end{gathered}$$

Yes, the food delivery company will deliver to the house because it is within the delivery radius.