If P and Q are the points of intersection of the circles x2 + y2 + 3x + 7y +2p – 5= 0 and x2 + y2 +2x + 2y – p2 = 0, then there is a circle passing through P, Q, and (1, 1) for -

let given circles be s= x2 + y2 + 3x + 7y +2p – 5= 0 and s1=x2 + y2 +2x + 2y – p2 = 0

Equation of required circle is s+s1=0

As it passes through (1,1) the value of=

if 7+2p=0 then it becomes the second circle,therefore it is true for all values of p.

Shaik Aasif

P and Q are the points of intersection of the line 
x/a + y/b=1 
 (a>O, b>O) 
with the x- and y-axes respectively. The distance PQ is 20 and the gradient of PQ is - 3. 
Find the values of a and b.