Tangents and the Radius!

We can apply the use of simple trigonometry here. We should first know that tangent to a circle is perpendicular to the radius of the circle at the point of contact of the circle and the tangent. Then, by observing the right angled triangle outside the circle having sides $3,4,5$ and the citcle, we get two equations ----

$\frac{3}{5}=\sin 37^{\circ}$ .....(i) and $\frac{r}{r+4}=\sin 37^{\circ}$ ........(ii)

From equations (i) and (ii) ------

$\frac{r}{r+4}=\frac{3}{5}$

$\implies 5r=3r+12 \implies 2r=12 \implies r=\boxed{6}$

Using similarity of triangles, we get: r/3 = (r + 4)/5 ==> r = 6.

I used Pythagoras Identity to solve it, and also used the fact that the lengths of two tangents on a same circle from the same point are same. In the bigger triangle, $(4 + r)^{2} = (5 + 3)^{2} + r^{2}$ Simplifying, $r = 6.$

using simmetry from small and large triangle, we get :

$\frac{r}{3}=\frac{r+4}{5}$

$5r = 3r + 12$

$2r = 12$

$r = 6$

i solved it in an entirely different way. ie, by solving the equation (5+x)^2=(4+2r)4 provided that the value of x=3, we get the value for x as 3 because length of two tangents intersecting outside the circle are equal.

the formula applied here is If a secant and a tangent of a circle are drawn from a point outside the circle, then the product of the lengths of the secant and its external segment equals the square of the length of the tangent segment.

No sine needed, based on power of a point to a circle (not sure how it is translated).

1. Take the right point of the 5 against the circle. Therefore the upper tangent is 5+3(also three)

2. Based on power of the point with the angle next to it we can say that 8 8=4 (4+2r) We get a simple equation where r = 6.

QED