$Find the length of the common chord of two intersecting circles.$

Let the two circles be \(C_1\) with radius \(r_1\) and center \(O_1\), and \(C_2\) with radius \(r_2\) and center \(O_2\). The distance between the centers is \(d\). The common chord has endpoints where the circles intersect.

Let the length of the chord be \(2l\). The perpendicular distance from the midpoint of the chord to the line joining the centers is given by:

$h = \sqrt{r_1^2 – l^2} = \sqrt{r_2^2 – l^2}$

From the triangle formed by \(O_1, O_2\) and the midpoint of the chord, we have:

$O_1M^2 + O_2M^2 = d^2$

Substituting, we get:

$\sqrt{r_1^2 – l^2} + \sqrt{r_2^2 – l^2} = d$

Solving for \(l\),

$l = \sqrt{r_1^2 – \left(\frac{d}{2}\right)^2} = \sqrt{r_2^2 – \left(\frac{d}{2}\right)^2}$

Thus, the length of the common chord is:

$2l = 2\sqrt{r_1^2 – \left(\frac{d}{2}\right)^2}$

Given two circles with equations:

$x^2 + y^2 = r_1^2$

$ (x – d)^2 + y^2 = r_2^2$

To find the intersections, equate the equations:

$x^2 + y^2 = (x – d)^2 + y^2$

Simplifying,

$x^2 = (x – d)^2$

Which gives:

$x^2 – (x^2 – 2dx + d^2) = 0$

$2dx = d^2$

$x = frac{d}{2}$

Substitute back to find y:

$r_1^2 – left(frac{d}{2}
ight)^2 = y^2$

$y = sqrt{r_1^2 – left(frac{d}{2}
ight)^2}$

The length of the common chord is:

$2y = 2sqrt{r_1^2 – left(frac{d}{2}
ight)^2}$

Let the radii be (r_1) and (r_2), and the distance between the centers be (d).

Using the formula for the chord length,

$2sqrt{r_1^2 – left(frac{d}{2}
ight)^2}$

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