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One of their common tangents is touching as shown in the figure.  Find the radius of the circle with center R.

May 25, 2020

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One of their common tangents is touching as shown in the figure.

Find the radius of the circle with center R.

$$\text{Let PR=p+r} \\ \text{Let PB=p-r} \\ \text{Let QR=q+r} \\ \text{Let QC=q-r} \\ \text{Let PQ=p+q} \\ \text{Let PA=p-q} \\ \text{Let DF=AQ} \\ \text{Let DF=DE+EF}$$

$$\begin{array}{|rcll|} \hline \mathbf{DE^2 +PB^2} &=& \mathbf{PR^2} \\ DE^2 + (p-r)^2 &=& (p+r)^2 \\ DE^2 &=& (p+r)^2 - (p-r)^2 \\ DE^2 &=& p^2+2pr+r^2-p^2+2pr-r^2 \\ DE^2 &=& 2*2pr \\ \mathbf{DE} &=& \mathbf{2\sqrt{pr} } \\ \\ \hline \mathbf{EF^2 +QC^2} &=& \mathbf{QR^2} \\ EF^2 + (q-r)^2 &=& (q+r)^2 \\ EF^2 &=& (q+r)^2 - (q-r)^2 \\ EF^2 &=& q^2+2qr+r^2-q^2+2qr-r^2 \\ EF^2 &=& 2*2qr \\ \mathbf{EF} &=& \mathbf{2\sqrt{qr} } \\ \\ \hline \mathbf{DF^2 + PA^2} &=& \mathbf{PQ^2} \\ DF^2 + (p-q)^2 &=& (p+q)^2 \\ DF^2 &=& (p+q)^2 - (p-q)^2 \\ DF^2 &=& p^2+2pq+q^2-p^2+2pq-q^2 \\ DF^2 &=& 2*2pq \\ \mathbf{DF} &=& \mathbf{2\sqrt{pq} } \\ \hline \end{array}$$

$$\begin{array}{|rcll|} \hline \mathbf{DF} &=& \mathbf{DE+EF} \\ 2\sqrt{pq} &=& 2\sqrt{pr}+2\sqrt{qr} \quad | \quad : 2 \\ \sqrt{pq} &=& \sqrt{pr}+\sqrt{qr} \\ \sqrt{p}\sqrt{q} &=& \sqrt{p}\sqrt{r}+\sqrt{q}\sqrt{r} \quad | \quad : \sqrt{r} \\\\ \dfrac{\sqrt{p}\sqrt{q}}{\sqrt{r}} &=& \sqrt{p} +\sqrt{q} \quad | \quad : \sqrt{p}\sqrt{q} \\\\ \dfrac{1}{\sqrt{r}} &=& \dfrac{\sqrt{p}}{\sqrt{p}\sqrt{q}} +\dfrac{\sqrt{q}}{\sqrt{p}\sqrt{q}} \\\\ \mathbf{\dfrac{1}{\sqrt{r}}} &=& \mathbf{\dfrac{1}{\sqrt{q}} +\dfrac{1}{\sqrt{p}}} \\ \hline \end{array}$$

$$\begin{array}{|rcll|} \hline \mathbf{\dfrac{1}{\sqrt{r}}} &=& \mathbf{\dfrac{1}{\sqrt{q}} +\dfrac{1}{\sqrt{p}}} \quad | \quad p=16,\ q=4 \\\\ \dfrac{1}{\sqrt{r}} &=& \dfrac{1}{\sqrt{4}} +\dfrac{1}{\sqrt{16}} \\\\ \dfrac{1}{\sqrt{r}} &=& \dfrac{1}{2} +\dfrac{1}{4} \\\\ \dfrac{1}{\sqrt{r}} &=& \dfrac{3}{4} \\\\ \sqrt{r} &=& \dfrac{4}{3} \quad | \quad \text{square both sides}\\\\ r &=& \dfrac{16}{9} \\\\ r &=& 1.\bar{7}\ \text{cm} \\ \hline \end{array}$$

The radius of the circle with center R is $$\mathbf{1.\bar{7}\ \text{cm}}$$

May 26, 2020