`x^2-10x+16`

The two circles x^2+ y^2=r^2 and x^2+y^2-10x +16=0 intersect at two distinct points. Then

The two circles x^2+ y^2=r^2 and x^2+y^2-10x +16=0 intersect at two distinct points. Then

Prove that the centres of the three circles : x^2 + y^2 -4x-6y-14=0, x^2 + y^2+ 2x+4y-5=0 and x^2 + y^2-10x -16y + 7 = 0 are collinear.

The number of integral value of x satistying sqrt( x^2+10 x-16)ltx-2 is

The number of integral value of x satistying sqrt( x^2+10 x-16)ltx-2 is

The two circles x^(2)+y^(2)=r^(2) and x^(2)+y^(2)-10x+16=0 intersect at two distinct points.Then

The circles x^2+y^2=r^2 and x^2+y^2-10x+16=0 intersect each other in distinct points if:

The circles : x^2+y^2 - 10x + 16 =0 and x^2+y^2 =a^2 : intersect at two distinct points if :