Find the radius $r$ of the semicircumference below

Question

In the figure, if $PF=3$, $OP=1$, calculate "$r$".

Answer by plan geometry

I try

$FM \cdot FN = FA\cdot FB$

$FA =4 - r$

$\therefore (4-r)\cdot(FB) = FM\cdot FN \implies (4-r)\cdot(4-r+2r) = FM\cdot FN$

$\therefore 16-r^2=FM\cdot FN$

Answer 1

Below is your diagram with a couple of lines, a point and several line lengths added:

In particular, $LF$ and $LO$ have been added, with $\lvert FL\rvert = a$. Since $LM$ and $LN$ are tangent to the circle, then $LO$ is a perpendicular bisector of $MN$, intersecting at $Q$, with $\lvert LQ\rvert = b$, $\lvert MQ\rvert = \lvert QN\rvert = c$ and $\lvert QO\rvert = d$. Finally, $\lvert FM\rvert = e$ and $\lvert LP\rvert = h$.

Since $\lvert FP\rvert = 3$, we have using the Pythagorean theorem with $\triangle FLP$ and $\triangle FLQ$ that

$$a^2 = 9 + h^2 = (e + c)^2 + b^2 \tag{1}\label{eq1A}$$

Also, since $\lvert PO\rvert = 1$, the Pythagorean theorem with $\triangle PLO$ gives

$$1 + h^2 = (b + d)^2 \tag{2}\label{eq2A}$$

Subtracting \eqref{eq2A} from \eqref{eq1A} gives

$$8 = (e^2 + 2ec + c^2 + b^2) - (b^2 + 2bd + d^2) = e^2 + 2ec + c^2 - 2bd - d^2 \tag{3}\label{eq3A}$$

We have $\angle LNQ + \angle NLQ = \angle LNQ + \angle QNO = 90^{\circ} \;\;\to\;\; \angle NLQ = \angle QNO$. Thus, $\triangle NQL \sim \triangle OQN$ so

$$\frac{\lvert LQ\rvert}{\lvert QN\rvert} = \frac{\lvert QN\rvert}{\lvert QO \rvert} \;\;\to\;\; \frac{b}{c} = \frac{c}{d} \;\;\to\;\; bd = c^2 \tag{4}\label{eq4A}$$

Using this in \eqref{eq3A} gives

$$8 = e^2 + 2ec - c^2 - d^2 \tag{5}\label{eq5A}$$

From your result, and my line lengths, we have

$$16 - r^2 = \lvert FM\rvert \cdot \lvert FN\rvert = e(e + 2c) = e^2 + 2ec \tag{6}\label{eq6A}$$

Using this, and the Pythagorean theorem with $\triangle OQN$ giving that $c^2 + d^2 = r^2$, in \eqref{eq5A} produces

$$8 = (16 - r^2) - r^2 \;\;\to\;\; 2r^2 = 8 \;\;\to\;\; r = 2 \tag{7}\label{eq7A}$$

Answer 2

Set up coordinate system, with origin at $O$, then we have $F(-4,0), P(-1,0)$.

Let angle $\angle NOB=\theta$, and radius is $r$. So we have $N(r\cos\theta, r\sin\theta)$, and the line equation for $NL$

$$y-r\sin\theta=-\frac{1}{\tan\theta}(x-r\cos\theta)$$

We get the coordinate for point $L$

$$L\left(-1, r\sin\theta-\frac{1}{\tan\theta}(-1-r\cos\theta)\right)\Rightarrow L\left(-1, \frac{r+\cos\theta}{\sin\theta}\right)$$

Slope for line $OL$

$$k_{OL}=-\frac{r+\cos\theta}{\sin\theta}$$

Slope for line $FN$

$$k_{FN}=\frac{r\sin\theta}{r\cos\theta+4}$$

$OL\perp FN\Rightarrow k_{OL}\cdot k_{FN}=-1$, so we get

$$\frac{r+\cos\theta}{\sin\theta}\cdot\frac{r\sin\theta}{r\cos\theta+4}=1$$ Simplify and we get $$\Rightarrow r^2\sin\theta=4\sin\theta$$

Cancel the $\sin\theta$ term, we get

$$r=2$$

Answer 3

The Points $F, A, P,B$ form one harmonic quadruple $\therefore \frac{FA}{AP} = \frac{FB}{PB} \implies \frac{4-R}{R-1}=\frac{4+R}{R+1}\\ 4R+4-R^2-R =4R-4+R^2-R \implies R^2=4 \therefore \boxed{R=2}$

(Answer by geobson)