Answer:
We can start by drawing a diagram of the situation. Let O be the center of both circles and let K be a point on the circumference of ⊙J such that JK = 12 cm. Let L be the point where JK intersects ⊙M, and let M be the point diametrically opposite L on ⊙M. Then, we have a right triangle JOK with legs JO = 32 cm and OK = 38 cm, and a right triangle LOM with legs LO = OM = r, where r is the radius of ⊙M. The hypotenuse of both triangles is the same and has length 64 cm.
We can use the Pythagorean theorem to find r. In the right triangle JOK, we have:
JO^2 + OK^2 = JK^2
32^2 + 38^2 = JK^2
JK = sqrt(32^2 + 38^2) ≈ 49.21 cm
In the right triangle LOM, we have:
LO^2 + OM^2 = LM^2
r^2 + r^2 = (2r)^2
2r^2 = LM^2
We know that LM = 2r + 12, since JK = 12 cm. Substituting this into the equation above, we get:
2r^2 = (2r + 12)^2
2r^2 = 4r^2 + 48r + 144
2r^2 - 4r^2 - 48r - 144 = 0
-r^2 - 24r - 72 = 0
r^2 + 24r + 72 = 0
We can solve for r using the quadratic formula:
r = (-24 ± sqrt(24^2 - 4*72)) / 2
r = (-24 ± sqrt(384)) / 2
r = -12 ± 4sqrt(6)
Since r is the radius of ⊙M, we want the positive value of r. Therefore:
r = -12 + 4sqrt(6) ≈ 4.03 cm
Finally, we can find LM:
LM = 2r + 12
LM = 2(4.03) + 12
LM ≈ 20.06 cm
Therefore, the length of LM is approximately 20.06 cm.