The line PC is a diameter d of the circle, where P is the tangent point on AB. Let AP=x. In the right-angled triangle APC, x^2+d^2=AC^2=9; similarly, in CPB, (5-x)^2+d^2=16, because we have two right-angled triangles with common side PC. If we subtract the first equation from the second to eliminate d^2, we get 5(5-2x)=7, 25-10x=7 and 10x=18, so x=1.8. d^2=9-x^2=9-3.24=5.76, so d=2.4. Because the angle in a semicircle is a right angle, CJP=PKC and JPKC is a rectangle with JK=PC=d=2.4.