Q.45 - 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 option (b) ${(x+1)}^2+{(y-2)}^2=4$

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 standard form of equation is ${(x-h)}^2+{(y-k)}^2=r^2 , where (h,k) is the center and r is the radius$

so here given equation ${(x-3)}^2+{(y+3)}^2=9$ is in the form of standard form of the equation of a circle and then the radius is 3 and center is (3,-3) , the graph is given below

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

${(x+1)}^2+{(y-2)}^2=4$

is the equation of a circle which is in the form of standard form of the equation of a circle , then radius is 2 and center is(-1, 2) here. Hence sketched graph of this equation is given below

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

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

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

Here given equation is ${(x+3)}^2+{(y-3)}^2=9$

  and compared to this  to standard form of a circle the  we get center is (-3, 3) and radius is 3. then the graph is given below 

Step 6. Matching the correct equation

The plotted graph of the equation for each option and the exact match of the equation of  the given question  graph is option (b)${(x+1)}^2+{(y-2)}^2=4$