One of their common tangents is touching as shown in the figure. Find the radius of the circle with center R.
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 \(\mathbf{1.\bar{7}\ \text{cm}}\)