Given two distinct intersecting circles, length of that chord of larger circle which is bisected by the smaller circle is equal to?

Solution 1:

Drop a perp from $C_1$ to $AB$ and extend. Then $D$ is the midpoint of $AB$. Drop a perp from $C_2$ to $C_1D$ extend.

If $AC = 4a$, $C_2E = BD = a$ and $C_1E = C_1D + C_2B$

$C_1D = \sqrt{1-a^2}, C_2B = \sqrt{2 - 4a^2}$

Using Pythagoras in $\triangle C_1EC_2$,

$ (\sqrt{1-a^2} + \sqrt{2 - 4a^2})^2 + a^2 = 4$

$ 3 - 4a^2 + 2 \sqrt{1-a^2} \sqrt{2 - 4a^2} = 4$

$ 4 (1-a^2) (2 - 4a^2) = (1 + 4a^2)^2$

$ 8 - 24a^2 + 16a^4 = 1 + 16 a^4 + 8 a^2$

$ \displaystyle 32a^2 = 7 \implies AC = 4a = \sqrt{\frac{7}{2}}$

Solution 2:

Hint :

Since $AC$ is bisected at $B$, $C_2B \perp AC$.

Hint 2 :

The circle with diameter $AC_2$ passes through $B$. Hence $AB$ is the radical axis of this circle and $S_1$.

Hint 3 :

The distance of any of the centers of three circles from the radical axis is readily calculable. Using some right triangles, length of $AC$ can then be determined.

Alternatively, one can go for geometric solution.

Denoting by $M$ midpoint of $AC_2$, $AC_1BM$ is a kite with four sides and one diagonal known (for reasons stated in Hint 2). Its other diagonal $AB$ can be found by application of Pythagoras theorem and doubled to give the length of $AC$.

Solution 3:

Yes there is a different path using trigonometry:

Let, with your coordinate system:

$$B=(\cos \alpha, \sin \alpha) \ \text{and} \ C=(2+\sqrt{2}\cos \beta, \sqrt{2}\sin \beta)$$

We just have to express that B is the midpoint of [AC] by writing that $$2B=A+C \ \iff \ \begin{cases}2\cos \alpha&=& \dfrac34+2+\sqrt{2}\cos\beta & (1a)\\2 \sin \alpha&=&\dfrac{\sqrt{7}}{4}+\sqrt{2}\sin \beta & (1b)\end{cases}$$

This gives you 2 equations in the 2 unknowns $\alpha$ and $\beta$.

Squaring and adding (1a) and (1b) gives :

$$4=(\dfrac{11}{4}+\sqrt{2}\cos\beta)^2+(\dfrac{\sqrt{7}}{4}+\sqrt{2}\sin \beta)^2$$

Expanding and using once more $\cos^2a +\sin^2 a=1$:

$$4=(\dfrac{11}{4})^2+2(\dfrac{11}{4})\sqrt{2}\cos\beta+(\dfrac{7}{16})^2+2\dfrac{\sqrt{7}}{4}\sqrt{2}\sin \beta+2$$

which is the very classical equation $A \cos a + B\sin a=C$