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

The lines quad p(p^(2)+1)x-y+q=0 and (p^(2)+1)^(2)x+(p^(2)+1)y+2q=0 are perpendicular to a common line for (1) no value of p(4) more than two values of p two values of p(4) more than two values of p

The radical centre of the three circles is at the origin. The equations of the two of the circles are x^(2)+y^(2)=1 and x^(2)+y^(2)+4x+4y-1=0 . If the third circle passes through the points (1, 1) and (-2, 1), and its radius can be expressed in the form of (p)/(q) , where p and q are relatively prime positive integers. Find the value of (p+q) .

If the circles x^(2)+y^(2) +5Kx+2y + K=0 and2(x^(2)+y^(2))+2Kx +3y-1=0, (KinR) , intersect at the points P and Q, then the line 4x + 5y - K = 0 passes through P and Q, for

If the point (3,4) lies inside and the point (-3,-4) lies outside the circle x ^(2) + y ^(2) - 7x +5y -p=0, then the set of all possible values of p is

If the mid-point of the line joining (3, 4) and (p, 7) is (x, y) and 2x + 2y + 1 = 0, then what will be the value of p?

If line x-2y-1=0 intersects parabola y^(2)=4x at P and Q, then find the point of intersection of normals at P and Q.