If P and Q are the points of intersection of the circles `x^2+""y^2+""3x""+""7y""+""2p""""5""=""0` and `x^2+""y^2+""2x""+""2y""""p^2=""0` , then there is a circle passing through P, Q and `(1,""1)` for (1) all values of p (2) all except one value of p (3) all except two values of p (4) exactly one value of p

Options

A. all values of p
B. all except one value of p
C. all except two values of p
D. exactly one value of p

Answer

A

Explanation

The equation of the circle passing through P and Q is
`(x^(2)+y^(2)+3x+7y+2p-5)+lambda(x^(2)+y^(2)+2x+2y-p^(2))=0...(i)` [Using : `S_(1)+lambdaS_(2)=0`]
It passes through (1,1) .
`:. (2p+7)+lambda(6-p^(2))=0rArr lambda=(2p+7)/(p^(2)-6)`
If `p=-(7)/(2)`, then `lambda=0` and equation (i) represents first circle.
If `p^(2)=6`, then equation (i) represents second circle. For all other values of p, equation represents some other circle. Hence, equation(i) represents a circle for all values of p.