Q. 59 - Precalculus Enhanced with Graphing Utilities

Question

Which of the following equations might have the graph shown? (More than one answer is possible.) 

$$\begin{gathered} \left(a\right) (x-2{)}^2+(y+3{)}^2=13 \\ \left(b\right) (x-2{)}^2+(y-2{)}^2=8 \\ \left(c\right) (x-2{)}^2+(y-3{)}^2=13 \\ \left(d\right) (x+2{)}^2+(y-2{)}^2=8 \\ \left(e\right) x^2+y^2-4x-9y=0 \\ \left(f\right) x^2+y^2+4x-2y=0 \\ \left(g\right) x^2+y^2-9x-4y=0 \\ \left(h\right) x^2+y^2-4x-4y=4 \end{gathered}$$

Step-by-Step Solution

Answer

Equations in (b),(c),(e) and (g) might have the graph shown. 

Step 1. Given information

We have been given a graph:

We have to tell which of the following equations might have the graph shown:

$$\begin{gathered} \left(a\right) (x-2{)}^2+(y+3{)}^2=13 \\ \left(b\right) (x-2{)}^2+(y-2{)}^2=8 \\ \left(c\right) (x-2{)}^2+(y-3{)}^2=13 \\ \left(d\right) (x+2{)}^2+(y-2{)}^2=8 \\ \left(e\right) x^2+y^2-4x-9y=0 \\ \left(f\right) x^2+y^2+4x-2y=0 \\ \left(g\right) x^2+y^2-9x-4y=0 \\ \left(h\right) x^2+y^2-4x-4y=4 \end{gathered}$$

Step 2. Analyze the graph

Given graph is a circle.

Origin $(0,0)$ should satisfy the equation of the circle and the center lies in the first quadrant.

Step 3. Find the equations that pass through the origin

$$\begin{gathered} \left(a\right) (x-2{)}^2+(y+3{)}^2=13 \\ \mathrm{Put} (0,0) \\ (0-2{)}^2+(0+3{)}^2=13 \\ (-2{)}^2+(+3{)}^2=13 \\ 4+9=13 \\ 13=13 \\ \left(b\right) (x-2{)}^2+(y-2{)}^2=8 \\ \mathrm{Put} (0,0) \\ (0-2{)}^2+(0-2{)}^2=8 \\ 4+4=8 \\ 8=8 \\ \left(c\right) (x-2{)}^2+(y-3{)}^2=13 \\ \mathrm{Put} (0,0) \\ (0-2{)}^2+(0-3{)}^2=13 \\ 4+9=13 \\ 13=13 \\ \left(d\right) (x+2{)}^2+(y-2{)}^2=8 \\ \mathrm{Put} (0,0) \\ (0+2{)}^2+(0-2{)}^2=8 \\ 4+4=8 \\ 8=8 \\ \left(e\right) x^2+y^2-4x-9y=0 \\ \mathrm{Put} (0,0) \\ \left(f\right) x^2+y^2+4x-2y=0 \\ \mathrm{Put} (0,0) \\ {0}^2+0^2+4\left(0\right)-2\left(0\right)=0 \\ 0=0 \\ \left(g\right) x^2+y^2-9x-4y=0 \\ \mathrm{Put} (0,0) \\ {0}^2+0^2-9\left(0\right)-4\left(0\right)=0 \\ 0=0 \\ \left(h\right) x^2+y^2-4x-4y=4 \\ \mathrm{Put} (0,0) \\ {0}^2+0^2-4\left(0\right)-4\left(0\right)=4 \\ 0\ne 4 \end{gathered}$$

All equations except (h) might have the graph. 

Step 4. Check the center of the given equations of the circle

$$\begin{gathered} \left(a\right) (x-2{)}^2+(y+3{)}^2=13 \\ Center=(2,-3) \\ \left(b\right) (x-2{)}^2+(y-2{)}^2=8 \\ Center=(2,2) \\ \left(c\right) (x-2{)}^2+(y-3{)}^2=13 \\ Center=(2,3) \\ \left(d\right) (x+2{)}^2+(y-2{)}^2=8 \\ Center=(-2,2) \\ \left(e\right) x^2+y^2-4x-9y=0 \\ {(x-2)}^2+{(y-\frac{9}{2})}^2=\frac{97}{4} \\ Center=(2,\frac{9}{2}) \\ \left(f\right) x^2+y^2+4x-2y=0 \\ {(x+2)}^2+{(y-1)}^2=5 \\ Center=(-2,1) \\ \left(g\right) x^2+y^2-9x-4y=0 \\ {(x-\frac{9}{2})}^2+{(y-2)}^2=\frac{97}{4} \\ Center=(\frac{9}{2},2) \end{gathered}$$

Therefore, (b),(c),(e) and (g) might have the graph shown.