Sine of angle between direct common tangents of two circles? This is a question I found: Given two circles intersecting orthogonally having the length of common chord 24/5 units the radius of one of the circles is 3 units then what is the sine of the angle between the direct common tangents? The answer given is 4√6/25. How do I approach this problem? I know that direct common tangents meet at a point that externally divides line joining centres of circles in ratio of their radii, but I am getting negative values when I try to solve this. How do I do this?
Using pythagoras theorem and similarity of triangles, you can easily find the other radius which is $4$ and distance between centres is $5$. 

Now let centre of circle with radius $r_1=3$ be $C_1$ and with radius $r_2=4$ be $C_2$. Let intersection of direct tangent and line through $C_1C_2$ be $P$. 



Using similarity,

$$\frac{PC_1}{3} = \frac{PC_1+C_1C_2}{4}= \frac{PC_1+5}{4}$$
This gives $PC_1 = 15$. So now let half of the required angle be $\alpha$, this is the angle between direct tangent and line through $C_1C_2$. 

$$\sin \alpha = \frac{r_1}{PC_1} = \frac{1}{5},\,\, \cos\alpha = \frac{2\sqrt{6}}{5}$$

So required answer is $\sin 2\alpha = \frac{4\sqrt{6}}{25}$
To make things simpler. There's this formula you can memorise or derive as you may like by similarity and Pythagoras. 

Alpha{angle between DCT}= 2sin^-1[|r2-r1|/ d] 
Where R1 and R2 are radii and d is distance between centres.

Here, sin a/2 comes out to be 1/5.
Thus sin a = 2 sin a/2 cos a/2

= 4✓6/25