Given three circles `x^2+y^2-16x+60=0`, `3x^2+3y^2-36x+81=0` and `x^2+y^2-16x-12y+84=0`. Find (1) the points from which the tangents are equal in length and (2) this length.

Given three circles x^(2)+y^(2)-16x+60=03x^(2)+3y^(2)-36x+81=0 and x^(2)+y^(2)-16x-12y+84=0. Find (1) the points from which the tangents are equal in length and (2) this length.

The point from which the tangents to the circles x^2 +y^2-8x + 40 = 0,5x^2+5y^2 -25x +80=0, and x^2 +y^2-8x + 16y + 160 = 0 are equal in length, is

The point of contact of the circles x^(2)+y^(2)-4x+6y-3=0 and x^(2)+y^(2)+16x+6y+37=0 is

If the lengths of tangents drawn to the circles x^(2) + y^(2) - 8x + 40 = 0 5x^(2) + 5y^(2) - 25x + 80 = 0 x^(2) + y^(2) - 8x + 16y + 160 = 0 From the point P are equal, then P is equal to

The point from which the tangents to the circle x^2 + y^2 - 4x - 6y - 16 = 0, 3x^2 + 3y^2 - 18x + 9y + 6 = 0 and x^2 + y^2 - 8x - 3y + 24 = 0 are equal in length is : (A) (2/3, 4/17) (B) (17/16, 4/15) (C) (17/16, 4/15) (D) (5/4, 2/3)