Verified answer

Without loss of generality, place the center of circle A at the origin and the

center of circle B at (6,0). Then the equations of the circles are

x^2 + y^2 = 25 and (x - 6)^2 + y^2 = 49.

Subtract the first of these equations from the second to find that

(x - 6)^2 - x^2 = 24 ----> x^2 - 12x + 36 - x^2 = 24 ---> 12x = 12 ----> x = 1.

At x = 1 we have y^2 = 25 - 1^2 = 24 ----> y = +/- sqrt(24) = +/- 2*sqrt(6).

So the length of the common chord will just be 2*(2*sqrt(6)) = 4*sqrt(6).

Let us construct the smallest circle at origo. x^2+y^2=25.

The bigger circle has centre(6,0), Then (x-6)^2+y^2=49 give this one.

When x^2+y^2=(x-6)^2+y^2. x=3 is the solution. This point give y1=4, y2=-4-

Distance between the "crossing" points for the circles is 2*4=8.

MN has length 8.

triangle ABM and ABN are equal as their sides are 5,6,7

the common chord MN is twice the altitude h wrt the base AB = 6

If S is the area of triangles ABM and ABN, then

h = 2S/AB = 2S/6 = S/3

the area of a triangle whose sides a, b, c are given can be found using Heron's formula

call p = (a + b + c)/2 the half perimeter of the triangle

S = √(p(p - a)(p - b)(p - c))

with our data

p = 9, a = 5, b = 6, c = 7

S = √(9(9 - 5 )(9 - 6)(9 - 7)) = √216 = √6³ = 6√6

h = S/3 = 6√6/3 = 2√6

therefore MN = 2h = 4√6

hope it is useful