(b) $P T$ is a direct common tangent of the circles drawn. $A B$ is a common chord which has been produced to meet the common tangent at $C$. Use the 'square of the tangents' result to prove that $C P=C T$.

Step 1 ：Given two circles with a common chord AB, which is extended to meet the common tangent PT at point C.

Step 2 ：We are asked to prove that CP = CT.

Step 3 ：Consider the two triangles formed by the points P, C, A and C, T, B.

Step 4 ：By the 'square of the tangents' result, we know that PA^2 = PC^2 + AC^2 and TB^2 = TC^2 + BC^2.

Step 5 ：Since AB is a common chord, we know that AC = BC and PA = TB.

Step 6 ：Therefore, we can equate the two equations and solve for PC and TC.

Step 7 ：By using the 'square of the tangents' result and the properties of the common chord, we can prove that CP = CT.

Step 8 ：\(\boxed{C P=C T}\)