Given that the equation of the circle

x^{2}+y^{2}-2x-4y-20=0

=>(x-1)^{2}+(y-2)^{2}=20+1^{2}+2^{2}=5^{2}

The coordinates of the center C of the circle = (1,2)

So radius of the circle r= 5 unit = CR=CQ

Given the coordinates of external point P = (16,7)we get

PC^{2 }=[(16-1)^{2}+(7-2)^{2}]=250

So PR^{2}=PQ^{2}= PC^{2} -r^{2}= 250-5^{2}=225

Hence PR=PQ= 15

Area of PCQ and PCR= 1/2 *15*5 sq unit

Hence area of PQCR = 75 sq unit