Answer:
the diameter of the largest circle is 2r = 56/3 cm.
Explanation:
Let the largest circle have center O and radius r. Join P, Q, R and O as shown in the figure below.
Since the circle with center Q and the circle with center R touch each other externally, the distance between their centers is the sum of their radii. Therefore,
QR = QM + MR
where M is the point of contact of the two circles. Similarly, since the circle with center P touches the circle with center Q externally, the distance between their centers is the sum of their radii. Therefore,
PQ = PM + MQ
Adding these two equations, we get
PQ + QR = PM + MQ + QM + MR
Substituting the given values, we get
10 + 12 = PM + MQ + 12 + 8
or PM + MQ = 2
Now consider the right-angled triangle PQO. Since PQ is the tangent to the circle with center P at point A, OA is perpendicular to PQ. Similarly, OQ is perpendicular to QR and OR is perpendicular to RP. Therefore, angles PAO, QBO and ROC are all right angles.
Let the perpendicular from O to PQ meet PQ at X. Then PX = OQ - PQ/2 = r - 5. Similarly, let the perpendicular from O to QR meet QR at Y. Then QY = OR - QR/2 = r - 6. Let the perpendicular from O to RP meet RP at Z. Then RZ = OP - PR/2 = r - 4.
Now consider the right-angled triangle OXY. Using the Pythagorean theorem, we get
OX^2 = OY^2 + XY^2
Substituting the values of OY and XY, we get
(r - 6)^2 = (r - 5)^2 + (r - 4)^2
Expanding and simplifying, we get
3r^2 - 56r + 191 = 0
Solving this quadratic equation, we get two solutions: r = 7 and r = 28/3. Since the radius of the largest circle cannot be less than the radius of the circle with center Q, which is 6, the only possible solution is r = 28/3.
Therefore, the diameter of the largest circle is 2r = 56/3 cm.