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

$$\begin{gathered} \left(a\right)(x-2{)}^2+y^2=3 \\ (b)(x+2{)}^2+y^2=3 \\ \left(c\right)x^2+(y-2{)}^2=3 \\ (d)(x+2{)}^2+y^2=4 \\ \left(e\right)x^2+y^2+10x+16=0 \\ \left(f\right)x^2+y^2+10x-2y=1 \\ \left(g\right)x^2+y^2+9x+10=0 \\ \left(h\right)x^2+y^2-9x-10=0 \end{gathered}$$

We have been given a graph:

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

$$\begin{gathered} \left(a\right)(x-2{)}^2+y^2=3 \\ (b)(x+2{)}^2+y^2=3 \\ \left(c\right)x^2+(y-2{)}^2=3 \\ (d)(x+2{)}^2+y^2=4 \\ \left(e\right)x^2+y^2+10x+16=0 \\ \left(f\right)x^2+y^2+10x-2y=1 \\ \left(g\right)x^2+y^2+9x+10=0 \\ \left(h\right)x^2+y^2-9x-10=0 \end{gathered}$$

Given graph is a circle.

The center of the circle has a negative x-coordinate.

Since y-coordinate of the center is zero, the equation of the circle should not have any term containing y only.

Equation (f) has a term with y, so we will not consider this equation.

$$\begin{gathered} \left(a\right)(x-2{)}^2+y^2=3 \\ \mathrm{Center}=(2,0\left) \\ \right(b)(x+2{)}^2+y^2=3 \\ \mathrm{Center}=(-2,0) \\ \left(c\right)x^2+(y-2{)}^2=3 \\ \mathrm{Center}=(0,2\left) \\ \right(d)(x+2{)}^2+y^2=4 \\ \mathrm{Center}=(-2,0) \\ \left(e\right)x^2+y^2+10x+16=0 \\ {(x+5)}^2+y^2=9 \\ \mathrm{Center}=(-5,0) \\ \left(g\right)x^2+y^2+9x+10=0 \\ {(x+\frac{9}{2})}^2+y^2=\frac{41}{4} \\ \mathrm{Center}=(-\frac{9}{2},0) \\ \left(h\right)x^2+y^2-9x-10=0 \\ {(x-\frac{9}{2})}^2+y^2=\frac{121}{4} \\ \mathrm{Center}=(\frac{9}{2},0) \end{gathered}$$

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