Draw a circle with centre `O` and any radius. Draw `A C\ a n d\ B D` two perpendicular diameters of the circle. Join `A B ,\ B C ,\ C D\ a n d\ D A` .

Draw circles with centres A,B and C each of radius 3 cm, such that each circle touches the other two circles.

Two circle intersect in A and B and AC and AD are respectively the diameters of the circles.Prove that C,B,D are collinear.

Two circle with centres A and B intersect at C and D. Prove that /_ACB=/_ADB

In Fig. 15.63, A B C D is a trapezium with A B D C and /_B C D=60o . If B F E C is a sector of a circle with centre C and A B=B C=7\ c m and D E=4\ c m , then find the area of the shaded region (U s e\ \ pi=(22)/7\ a n d\ sqrt(3)=1. 732) .

A B C is a triangle with B as right angle, A C=5\ c m\ a n d\ A B=4\ c mdot A circle is drawn with A as centre and A C as radius. The length of the chord of this circle passing through C\ a n d\ B is 3 cm (b) 4 cm (c) 5 cm (d) 6 cm

A B\ a n d\ C D are two parallel chords of a circle with centre O such that A B=6\ c m\ a n d\ C D=12\ c mdot The chords are on the same side of the centre and the distance between them is 3 cm. The radius of the circle is 6\ c m (b) 5\ sqrt(2)\ c m (c) 7\ c m (d) 3\ sqrt(5)\ c m

In Figure, B M a n d D N are both perpendiculars to the segments A C a n d B M=D N . Prove that A C bisects B D

Construct a regular pentagon in a circle of radius 4 cm. the following sentences are the stepts involved in the following construction. Arrange them in sequential order from first to last. (A) With D as and AB as radius draw an arc which cuts the circle with radius point C. (B) With A as centre and AB as radius draw an arc whch cuts the circle at the point C. (C) Consutruct a circle with radius 4 cm. (D) Join AB. (E) Draw two radii overline(OA) and overline(OB) such that angle AOB =72^@ . (F) Join AE, ED, DC and CB. (G) With E as centre and AB as radius draw an arc which cuts the circle at the point D.

A B C D is a cyclic quadrilateral whose diagonals A C\ a n d\ B D intersect at Pdot If A B=D C , prove that: P A B\ ~= P D C P A=P D\ a n d\ P C=P B (iii) A D B C