Q.44 - Precalculus Enhanced with Graphing Utilities

Question

match each graph with the correct equation.$$\begin{gathered} \left(a\right) {(x-3)}^2+{(y+3)}^2= 9, \left(b\right) {(x+1)}^2+{(y-2)}^2=4 , \\ \left(c\right) {(x-1)}^2+{(y+2)}^2=4 , \left(d\right) {(x+3)}^2+{(y-3)}^2=9 \end{gathered}$$

Step-by-Step Solution

Answer

The matched equation of the given graph is option (d) ${\left(x+3\right)}^2+{(y-3)}^2=9$

Step 1. Given information

The graph of the circle is given.

 we need to match the graph with the correct equation.

Step 2 . Plot the graph of option (a) ( x - 3 ) 2 + ( y + 3 ) 2 = 9

The given equation of a circle is ${(x-3)}^2+{(y+3)}^2=9$   which is in the standard form of the equation of a circle

Thus here center is (h, k ) is (3,-3) and radius is 3. Then the graph of the equation ${(x-3)}^2+{(y+3)}^2=9.$  is given below 

Step 3. Plot the graph of option ( b )   ( x + 1 ) 2 + ( y - 2 ) 2 = 4

Here   equation of the circle is ${(x+1)}^2+{(y-2)}^2=4$.while comparing this equation to a standard form of the equation of a circle , we get center is (-1,2) and radius is 4. Using this we can plot the graph of the circle ,the plotted graph is given below

Step 4. Sketch the graph of option (c) x - 1 2 + ( y + 2 ) 2 = 4

we know that the standard form of the equation of a circle is ${(x-h)}^2+{(y-k)}^2=r^2$.

The given equation is ${(x-1)}^2+{(y+2)}^2=4$, when comparing given equation to the standard equation, then the center becomes (1,-2) and radius will become  2. Thus the graph of the circle is given below

Step 5. Sketch the graph of option (d) ( x + 3 ) 2 + ( y - 3 ) 2 = 9

 when we compared  ${(x+3)}^2+{(y-3)}^2=9$ to the standard form of the equation of circle we get center is (-3,3) and radius is 3.Now we can plot the graph , that is given below 

step 6. Matching the correct equation

The sketched graph of the equation for each option and the correct match of the equation of  the given question  graph is option (d)${(x+3)}^2+{(y-3)}^2=9$