Chapter 13.7, Problem 25PSC

To determine

To find the length of the common internal agent.

Explanation of Solution

Given:

The circles are represented by ${(x+9)}^2+{(y-4)}^2=52$ and ${(x-12)}^2+{(y-3)}^2=13$

Calculation:

The coordinates of center are $(-9,4)$ and $(12,3)$

  

  $\begin{array}{l}PQ=\sqrt{12-(-9)+{(3-4)}^2}=\sqrt{442}\\ \frac{PS}{SQ}=\frac{\sqrt{52}}{\sqrt{13}}=2\\ PS+SQ=\sqrt{442}\\ SQ=\frac{\sqrt{442}}{3},PS=\frac{2\sqrt{442}}{3}\\ RS=\sqrt{P{S}^2-P{R}^2}ST=\sqrt{S{Q}^2-T{Q}^2}\\ =\sqrt{\frac{4\times 442}{9}-52}=\frac{10\sqrt{13}}{3}=\sqrt{\frac{1}{9}\times 442-13}=\frac{5\sqrt{13}}{3}\end{array}$

The length of the common internal tangent is

  $RT=RS+ST=5\sqrt{13}$

To determine

To find the length of the common external agent.

Explanation of Solution

Given:

The circles are represented by ${(x+9)}^2+{(y-4)}^2=52$ and ${(x-12)}^2+{(y-3)}^2=13$

Calculation:

The length of external tangent is

  $\sqrt{442-13}=\sqrt{429}=20.712$