Chapter 9.3, Problem 49PPS

(a)

To determine

To Find: The equation of a circle concentric with the circle at the right with the radius $4units$ greater.

Answer to Problem 49PPS

  ${\left(x+1\right)}^2+{\left(y-4\right)}^2=36+16\sqrt{5}$

Explanation of Solution

Given: $AB$ is the diameter of the circle.

  

Equation of the circle

  ${(x-h)}^2+{(y-k)}^2=r^2$

Here

  $\begin{array}{l}A=(-5,2)\\ B(3,6)\end{array}$

Thus

Find the center using the Mid-Point formula

  $(h,k)=\left(\frac{x_1+x_2}{2},\frac{y_1+y_2}{2}\right)$

  $\begin{array}{l}{x}_1=-5\\ y_1=2\\ x_2=3\\ y_2=6\end{array}$

Now

  $\begin{array}{l}(h,k)=\left(\frac{x_1+x_2}{2},\frac{y_1+y_2}{2}\right)\\ (h,k)=\left(\frac{-5+3}{2},\frac{2+6}{2}\right)\\ (h,k)=\left(-1,4\right)\end{array}$

Now

Find the radius using the distance formula

  $r=\sqrt{{(x_2-x_1)}^2+{(y_2-y_1)}^2}$

Here

  $\begin{array}{l}{x}_1=-12\\ y_1=16\\ x_2=-1\\ y_2=4\end{array}$

  $\begin{array}{l}d=\sqrt{{(x_2-x_1)}^2+{(y_2-y_1)}^2}\\ d=\sqrt{{\left(3+5\right)}^2+{\left(6-2\right)}^2}\\ d=\sqrt{64+16}\\ d=\sqrt{80}\\ d=4\sqrt{5}\\ r=\frac{4\sqrt{5}}{2}\\ r=2\sqrt{5}\end{array}$

Thus

Equation of the circle is

  $\begin{array}{l}{\left(x+1\right)}^2+{\left(y-4\right)}^2={\left(2\sqrt{5}\right)}^2\\ {\left(x+1\right)}^2+{\left(y-4\right)}^2=80\end{array}$

Now the new center with the radius $4units$ greater.

  $\begin{array}{l}r=2\sqrt{5}+4\\ (h,k)=(-1,4)\end{array}$

Equation of the new circle is

  $\begin{array}{l}{\left(x+1\right)}^2+{\left(y-4\right)}^2={\left(2\sqrt{5}+4\right)}^2\\ {\left(x+1\right)}^2+{\left(y-4\right)}^2=36+16\sqrt{5}\end{array}$

Hence, ${\left(x+1\right)}^2+{\left(y-4\right)}^2=36+16\sqrt{5}$

(b)

To determine

To Find: The equation of a circle concentric with the circle at the right with the radius $2units$ less.

Answer to Problem 49PPS

  ${\left(x+1\right)}^2+{\left(y-4\right)}^2=24-8\sqrt{5}$

Explanation of Solution

Given: $AB$ is the diameter of the circle.

  

Equation of the circle is

  $\begin{array}{l}{\left(x+1\right)}^2+{\left(y-4\right)}^2={\left(2\sqrt{5}\right)}^2\\ {\left(x+1\right)}^2+{\left(y-4\right)}^2=80\end{array}$

Now the new center with the radius $2units$ less

  $\begin{array}{l}r=2\sqrt{5}-2\\ (h,k)=(-1,4)\end{array}$

Equation of the new circle is

  $\begin{array}{l}{\left(x+1\right)}^2+{\left(y-4\right)}^2={\left(2\sqrt{5}-2\right)}^2\\ {\left(x+1\right)}^2+{\left(y-4\right)}^2=24-8\sqrt{5}\end{array}$

Hence, ${\left(x+1\right)}^2+{\left(y-4\right)}^2=24-8\sqrt{5}$

(c)

To determine

To Graph: The circles in part (a) and (b)

Answer to Problem 49PPS

  

Explanation of Solution

Given: $AB$ is the diameter of the circle.

  

Equation of the circle is

  ${\left(x+1\right)}^2+{\left(y-4\right)}^2=80$

Equation of the two concentric circle is

  $\begin{array}{l}{\left(x+1\right)}^2+{\left(y-4\right)}^2=36+16\sqrt{5}\\ {\left(x+1\right)}^2+{\left(y-4\right)}^2=24-8\sqrt{5}\end{array}$

Thus

The graph