Chapter 13.6, Problem 7PSA

To determine

To find: The coordinate A, B, C and D of the given circle.

Answer to Problem 7PSA

The coordinate of A is $\left(-6,3\right)$ , B is $\left(8,3\right)$ , C is $\left(1,10\right)$ and D is $\left(1,-4\right)$ .

Explanation of Solution

Given:

The radius of the circle is 7.

The line AB is the horizontal diameter and CD is the vertical diameter.

The given diagram is shown in Figure 1

  

Figure 1

Calculation:

Consider the given equation is,

  ${\left(x-3\right)}^2+{\left(y+1\right)}^2=0$

Consider the formula for the equation of the circle is of the form as,

  ${\left(x-h\right)}^2+{\left(y-k\right)}^2=r^2$

Substitute $\left(1,3\right)$ in the above equation as,

  ${\left(x-1\right)}^2+{\left(y-3\right)}^2=7^2$

As the y coordinate of the point A and B is 3 then,

  $\begin{array}{l}{\left(x-1\right)}^2+{\left(3-3\right)}^2=7^2\\ \left(x-1\right)=\pm 7\\ x=8,-6\end{array}$

Thus, the coordinate of A is $\left(-6,3\right)$ and B is $\left(8,3\right)$ .

As the y coordinate of the point C and D is 1 then,

  $\begin{array}{l}{\left(1-1\right)}^2+{\left(y-3\right)}^2=7^2\\ {\left(y-3\right)}^2=7^2\\ \left(y-3\right)=\pm 7\\ y=10,-4\end{array}$

Thus, the coordinate of C is $\left(1,10\right)$ and D is $\left(1,-4\right)$ .

Thus, the coordinate of A is $\left(-6,3\right)$ , B is $\left(8,3\right)$ , C is $\left(1,10\right)$ and D is $\left(1,-4\right)$ .