Find radius of circle in the following circle.

Question

Let we have two tangent lines from point $A$ to circle.

Find radius of circle in the following circle:

I draw two radiuses of circle to the two tangent lines from point $A$ to circle so we have two triangles with angels $90$ but i don't know how we can use other information from question.

Answer 1

Let $c=AB$, $b=AC$ and $a=BC$ be tangent to the circle with $O$, $D$, $E$ and $F$, respectively

then $OD=OE=OF=r_a$

$\text{area of} \triangle ABC=\text{area of} \triangle AOB + \text{area of} \triangle AOC - \text{area of} \triangle BOC $

By Heron's formula we can compute the area of $\triangle ABC = \sqrt{u(u-a)(u-b)(u-c)}= \frac{b+c-a}{2}\times r_a = r_a (u-a)$ where $a=7,b=8,c=9$ and $2u=a+b+c$

$\sqrt{12(12-7)(12-8)(12-9)}=r_a (12-7)$

$r_a=\frac{12\sqrt5}{5}$

Answer 2

Let $O$ be the center of the circle, the points $D$ and $E$ be the intersection points of the lines $AC$ and $AB$ with the circle, respectively, point $F$ the intersection between $AO$ and $DE$, and point $P$ the intersection between $BC$ and the circle.

We have $PC = CD$ and $PB = BE$, since these are tangent segments drawn from the same point; so, since $PC + PB = 7$, then $CD + BE = 7$. But we also know that $CD + 8 = BE + 9$, so $CD = 4$, $BE = 3$ and $AD = AE = 12$. By the law os cosines, in $\triangle ABC$, we obtain $7^2 = 8^2 + 9^2 - 2\cdot 8 \cdot 9 \cdot \cos{B\hat{A}C}$, thus $\cos{B\hat{A}C} = \dfrac{2}{3}$. Then, in $\triangle AED$, we have $ED^2 = 288 - 2\cdot 144 \cdot \dfrac{2}{3}$ by the law of cosines, and $ED = 4\sqrt{6}$. Applying the pythagorean theorem in $\triangle AFD$:

$\left (\dfrac{4\sqrt{6}}{2}\right)^2 + AF^2 = 144$

$24 + AF^2 = 144$

$AF = 2\sqrt{30}$

Now, note that $\triangle AFD \sim \triangle AOD$, so $\dfrac{12}{OD} = \dfrac{2\sqrt{30}}{2\sqrt{6}}$, and hence $OD = \dfrac{12\sqrt{5}}{5}$.

Answer 3

Extend sides AB and AC. Draw bisector of $\angle BAC$. The radii of inscribed circle and the one it's radius to be found are on this line. Draw these circles. Draw a tangent on bigger circle parallel with BC.Name it's intersection with extension of AB as D and with extension of AC as E. Triangles ABC and ADE are similar, so the radii of circles is proportional to the altitudes of triangles and we may write:

$\frac {r_2}{r_1}=\frac {h_2}{h_1}$

Where $r_1$ is incircle radius, $r_2$ is required radius, $ h_1$ is the altitude of triangle ABC and $h_2$is the altitude of triangle ADE, we have:

$h_2=h_1+2r_1$

In triangle ABC we can find half perimeter as $p=(9+8+7)/2=12$ and by Heron's formula the are is $s=12\sqrt 5$ and $h_1=2\frac s{BC}$. Also we have :

$r_1=\frac s p=\frac{12\sqrt5}{12}=\sqrt 5$

$\frac {r_2}{r_1}=\frac{h_2}{h_1}\Rightarrow \frac{h_1+2r_2}{h_1}=\frac{r_2}{r_1}$

which finally gives:

$$r_2=\frac{r_1h_1}{h_1-2r_1}$$

$h_1=2\frac{12\sqrt5}7=\frac {24\sqrt 5}7$

putting values we finally get:

$$r_2=\frac{\sqrt5 \times 24\sqrt5}{7(\frac{24\sqrt 5}7-2\sqrt 5)}=\frac{12\sqrt 5}5$$