Sine of angle between direct common tangents of two circles?

Question

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?

Answer 1

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 $\sin 2\alpha = \frac{4\sqrt{6}}{25}$

Answer 2

Say one circle has equation $x^2+y^2=3^2$ and the other has $(x-5)^2+y^2=4^2.$ (In comments you say you've already figured on that the distance between centers is $5$ and the radius of the other circle is $4.$)

Let the equation of one of the common tangent lines be $y = m(x-x_0);$ the other must then be $y=-m(x-x_0).$ Then the line through the point of tangency with the smaller circle and the center of the smaller circle has equation $y=\dfrac{-1}mx$ and the one for the larger circle is $y=\dfrac{-1} m(x-5).$ Let $(x_1,y_1)$ be the point of tangency with the smaller circle and $(x_2,y_2)$ for the larger. Then we have \begin{align} y_1 & = m(x_1-x_0) \\ y_1 & = (-1/m)x_1 \\ & y_1^2+x_1^2 = 3^2 \\ y_2 & = m(x_2-x_0) \\ y_2 & = (-1/m)(x_2-5) \\ & y_2^2+x_2^2 = 4^2 \end{align}

You should be able to solve that system for $x_0,x_1,x_2,y_1,y_2,m.$

The sine of half the angle that you seek is $3/x_0,$ since you have a right triangle with hypotenuse of length $x_0$ and opposite side of length $3.$ The cosine can be found once you have the sine, and then the double-angle formula for the sine can be used.

Answer 3

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