Let ABCD be a square of side length 2 units. C2 is the circle through vertices A, B, C, D and C1 is the circle touching all the sides of the square ABCD. L is a line through A. . If P is a point on C1 and Q in another point on C2, then (PA^2+ PB^2+ PC^2+PD^2)/ (QA^2+ QB^2+ QC^2+ QD^2) is equal to (A) 0.75 (B) 1.25 (C) 1 (D) 0.5

Let ABCD be a square of side length 2 units. C_(2) is the fircle through the vertices A, B, C, D and C_(1) is the circle touching all the of the square ABCD. L is a lien through vertex A. A circle touches the line L and the circle C_(1) externally such that both the circles are on the same side of the line L. The locus of the centre of the circle is

Show that the points A (5, 6), B(1,5), C(2, 1) and D(6, 2) are the vertices of a square ABCD.

ABCD is a square of side length 2 units. C_(1) is the circle touching all the sides of the square ABCD and C_(2) is the circumcircle of square ABCD. L is a fixed line in the same plane and R is fixed point. If a circle is such that it touches the line L and the circle C_(1) externally, such that both the circles are on the same side of the line, then the locus of centre of the circle is

ABCD is a square of side length 2 units. C_(1) is the circle touching all the sides of the square ABCD and C_(2) is the circumcircle of square ABCD. L is a fixed line in the same plane and R is fixed point. A line L' through a is drawn parallel to BD. Point S moves scuh that its distances from the line BD and the vertex A are equal. If loucs S cuts L' at T_(2)andT_(3) and AC at T_(1) , then area of DeltaT_(1)T_(2)T_(3) is

If P be any point in the plane of square ABCD, prove that PA^(2)+PC^(2)=PB^(2)+PD^(2)

Show that the points A (5,6) B (1,5) and C (2,1) and D (6,2) are the vertices of a square ABCD.

Show that the points A(3,5),B(6,0), C(1,-3) and D(-2,2) are the vertices of a square ABCD.

A circle is inscribed into a rhombous ABCD with one angle 60. The distance from the centre of the circle to the nearest vertex is equal to 1. If P is any point of the circle then |PA|^2+|PB|^2+|PC|^2+|PD|^2 is equal to:

Consider one sides AB of a square ABCD in order on line y=2x-17 , and other two vertices C, D on y=x^(2) The maximum possible area of square ABCD is

Let ABCD be a square with sides of unit lenght. Points E and F are taken on sides AB and AD, respectively,so that AE=AF . Let P be a point inside the squre ABCD. The value of (PA)^2-(PB)^2+(PC)^2-(PD)^2 is equal to