smaller circle tangent to larger circle, ray PR passes thru

[judocallin]

Good Day! Please share your knowledge

In the figure, the smaller circle is tangent to the larger circle and ray PR passes through the centers of both circles. If radius of the smaller circle is a and the radius of the larger circle is b, show that sin?= ((b-a)/(b+a)). then using the identity ((sin)^(2))? +((cos)^(2))? =1 show that cos?= ((2sqrt(ab))/(a+b)).

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My Idea: I think it involves area but I just don't know how

Hoping for replies..Thanks

 

[soroban]

Re: Need Help

Hello, judocallin!

\(\displaystyle \text{In the figure, the smaller circle is tangent to the larger circle}\)
. . \(\displaystyle \text{and ray }PR\text{ passes through the centers of both circles.}\)

\(\displaystyle \text{(a) If radius of the smaller circle is }a \text{ and the radius of the larger circle is }b,\)
. . \(\displaystyle \text{show that: }\:\sin\theta \:=\:\frac{b-a}{b+a}\)

\(\displaystyle \text{(b) Then, using the identity }\sin^2\!\theta +\cos^2\!\theta \:=\:1,\)
. . \(\displaystyle \text{show that: }\:\cos\theta\:=\:\frac{2\sqrt{ab}}{a+b}\)

                               R    * * *
                                *           *
                  Q       *   *   \           *
                    *   *    *     \b          *
              *  *  \a     *        \
        * @     *    \      *        \          *
P * - - - - - - * - - * - - * - - - - * - - - - *
         x         a  S  a       b    T

The centers of the circles are S and T.
The ray from P is tangent to the circles at Q and R.
\(\displaystyle SQ = a,\;TR = b,\;\angle QPS = \theta\)

\(\displaystyle \text{In right triangle }SQP\!:\;\;\sin\theta \:=\:\frac{a}{x+a}\) .[1]

\(\displaystyle \text{In right triangle }TRP\!:\;\;\sin\theta \:=\:\frac{b}{x+2a+b}\) .[2]


Equate [1] and [2]:

.\(\displaystyle \frac{a}{x+a} \:=\:\frac{b}{x+2a+b} \quad\Rightarrow\quad x \:=\:\frac{2a^2}{b-a}\)


Substitute into [1]:

. . \(\displaystyle \sin\theta \:=\:\frac{a}{\frac{2a^2}{b-a} + a} \:=\:\frac{a(b-a)}{2a^2 + ab - a^2} \:=\:\frac{a(b-a)}{ab+a^2}\:=\:\frac{a(b-a)}{a(b+a)}\)

\(\displaystyle \text{(a) Therefore: }\;\boxed{\sin\theta \:=\:\frac{b-a}{b+a}}\)



\(\displaystyle \text{We have: }\;\sin^2\!\theta + \cos^2\!\theta \:=\:1\)

\(\displaystyle \text{Hence: }\:\left(\frac{b-a}{b+a}\right)^2 + \cos^2\!\theta \:=\:1\)

. . \(\displaystyle \cos^2\!\theta \:=\:1 - \frac{(b-a)^2}{(b+a)^2} \;=\;\frac{(b+a)^2-(b-a)^2}{(b+a)^2} \:=\:\frac{(b^2 + 2ab + a^2) - (b^2 - 2ab + b^2)}{(b+a)^2}\)

. . \(\displaystyle \cos^2\!\theta \:=\:\frac{4ab}{(b+a)^2} \quad\Rightarrow\quad \cos\theta \:=\:\sqrt{\frac{4ab}{(b+a)^2}}\)

\(\displaystyle \text{(b) Therefore: }\;\boxed{\cos\theta \;=\;\frac{2\sqrt{ab}}{a+b}}\)