From the point P(16,7) tangents PQ and PR are drawn to the circle :

Question

x²+ y² -2x-4y-20=0 . If C be the centre of the circle , then the area of the quadrilateral PQCR is:

• 450 sq. unit
• 15 sq. unit
• 50 sq. unit
• 75 sq. unit
• 175 sq. unit

Answer 1

Given that the equation of the circle

x²+y²-2x-4y-20=0

=>(x-1)²+(y-2)²=20+1²+2²=5²

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

PC² =[(16-1)²+(7-2)²]=250

So PR²=PQ²= PC² -r²= 250-5²=225

Hence PR=PQ= 15

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

Hence area of PQCR = 75 sq unit