Given two distinct intersecting circles, length of that chord of larger circle which is bisected by the smaller circle is equal to? Two circles whose centres lie on the x axis, whose radii are  $\sqrt2cm$ and $1cm$ and whose centres are 2 cm apart intersect at a point A.The chord AC of the larger circle cuts the smaller circle at a point B and is bisected by that point. What is the length of chord AC?
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My attempt :
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As shown in the diagram above I started by assuming the centre of smaller circle S1 (Of radius 1) to be origin and the centre of the larger circle S2 (Of radius $\sqrt2$ ) to be $(2,0)$ as the centres are separated by 2 units.
I then solved :
S1 : $ x^2 + y^2 = 1 $ ; and
S2 : $ (x-2)^2 + y^2 = 2 $ to obtain $A(\frac{3}4 , \frac{\sqrt7}4)$
I noted the following equations:

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*Since B is given to be mid point of chord AC:

XB = $\frac{X_A + X_C}2$ ; $Y_B = \frac{Y_A +Y_C}2$


*Since C lies on $S_2$ :

$ (X_C-2)^2 + Y_C^2 = 2 $


*The distance of B from origin is 1 unit :

$(X_B-0)^2 + (Y_B-0)^2 = 1$
$(\frac{X_A + X_C}2 - 0)^2$ +$(\frac{Y_A +Y_C}2 - 0)^2 = 1$
$(\frac{\frac{3}4 + X_C}2 - 0)^2$ +$(\frac{\frac{\sqrt7}4 +Y_C}2 - 0)^2 = 1$
This equation and equation generated in point 2 together are two equations in two variables and i should be able to solve them to get the co-ordinate of C. This however is proving to be cumbersome.
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Is there a better way to avoid this approach.
 A: 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}}$
A: 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 $AC$.

A: 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 $A \cos a + B\sin a=C$
