Given a circle of radius `r`. Tangents are drawn from point `A and B` lying on one of its diameters which meet at a point `P` lying on another diameter perpendicular to the other diameter. The minimum area of the triangle `PAB` is : (A) `r^2` (B) `2r^2` (C) `pir^2` (D) `r^2/2`

The total surface area of a hemisphere of radius r is pir^2 (b) 2pir^2 (c) 3pir^2 (d) 4pir^2

Two circles are drawn with sides A B ,A C of a triangle A B C as diameters. The circles intersect at a point D . Prove that D lies on B C .

From a fixed point A on the circumference of a circle of radius r , the perpendicular A Y falls on the tangent at Pdot Find the maximum area of triangle A P Ydot

Tangents are drawn to the circle x^2+y^2=50 from a point "P lying on the x-axis. These tangents meet the y-axis at points 'P_1,' and 'P_2 . Possible co-ordinates of 'P' so that area of triangle PP_1P_2 is minimum is/are -

The moment of inertia of a solid sphere of radius R about its diameter is same as that of a disc of radius 2R about its diameter. The ratio of their masses is

Let PQ and RS be tangents at the extremities of the diameter PR of a circle of radius r. If PS and RQ intersect at a point X on the circumference of the circle, then 2r equals

A circle of radius r cm has a diameter of length (a) r cm (b) 2r cm (c) 4r cm (d) r/2c m

Tangent is drawn at any point (p ,q) on the parabola y^2=4a x .Tangents are drawn from any point on this tangant to the circle x^2+y^2=a^2 , such that the chords of contact pass through a fixed point (r , s) . Then p ,q ,r and s can hold the relation (A) r^2q=4p^2s (B) r q^2=4p s^2 (C) r q^2=-4p s^2 (D) r^2q=-4p^2s

Let a and b represent the lengths of a right triangles legs. If d is the diameter of a circle inscribed into the triangle, and D is the diameter of a circle circumscribed on the triangle, the d+D equals. (a) a+b (b) 2(a+b) (c) 1/2(a+b) (d) sqrt(a^2+b^2)

If the circumference and the area of a circle are numerically equal, then diameter of the circle is (a) pi/2 (b) 2pi (c) 2 (d) 4