Q.46 - Precalculus Enhanced with Graphing Utilities

Question

match each graph with 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 correctly matched equation of the given question is an option (a)${(x-3)}^2+{(y+3)}^2=9$

Step 1. Given information

The graph of the circle is given.

 we have to match the graph with the correct equation.

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

 Here ${(x-3)}^2+{(y+3)}^2=9$ is the given equation of a circle, then the center is(3,-3) and the radius is 3 since given this equation compared to the standard form of equation of a circle.

Then the plotted graph is given below

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

The equation ${(x+1)}^2+{(y-3)}^2=4$ is similar to a standard form of the equation of a circle then center(-1,2) and radius is 2, thus the graph is given below 

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

 when we compared the given equation to  the standard for of the equation of a circle  , we get radius is 2 and center is(1,-2). Hence the graph of the equation given below

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

According to the given question ${(x+3)}^2+{(y-3)}^2=9$ , the  center is  

(-3, 3) and radius is 3 since the given equation is compared to the standard form of the equation of a circle. Hence the graph based on the given equation is below

Step 6. Matching the correct equation

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