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Welcome Part 1Part 2 20-Dec Dec-2015 Circle

1.Revision 2.Theorems 6 to 12 3.Practice Examples 4.Question and Answer

1.Center 2.Radius, Diameter – measure as well as line segment 3.Secant, Chord 4.Circumference, Area 5.Sector, Segment 6.Tangent – How many Inside the circle At the circle Outside the circle Basics

A Tangent at any point of a circle is perpendicular to the radius thru the point of contact Theorem 1 Proof: by contradiction – prove that we can find another point Q which is on the line and the circle

Converse 1.A line perpendicular to a radius of a circle at the outer end is tangent to the circle Theorem 2

1.The lengths of the two tangent segments to a circle from an external point are equal Theorem 3 Proof: prove that ∆ OAP and ∆ OBP are congruent by right angle ∆s side hypotenuse test

Touching circles – 5 possibilities 1.If two circles are touching then the common point lies on the line joining their centres Theorem 4 Proof: by contradiction – Similar to theorem 1. Prove that we can find another point Q which is on the line and the circle

Minor and Major Arcs, Inscribed angle - Angle within arc Intercepted Arc – arc within angle Angle subtended by an arc – the point is typically on the major arc The measure of an angle subtended by an arc is half the angle subtended by the same arc at the centre. Corollaries: 1.For Diameter, angle is perpendicular 2.Angles subtended by same arc and congruent Theorem 5 Proof: Using rule of isosceles ∆ and remote internal angle theorem

Tangent Secant Theorem Theorem 8

Tangent Secant Theorem If an angle with its vertex on the circle, whose one side touches the circle and other intersects the circle at two points (chord), then the angle is half the intercepted arc. ∠ ABC = m(arc AEB) Theorem 8 Proof: ∠ ABC + ∠ OBA = 90 0 ∠ OAB = ∠ OBA ∴ ∠ ABC + ∠ OAB = 90 0 and ∆ AOB sum of all angles is ∠ AOB + ∠ OAB + ∠ OBA = 180 0

The angle formed by a secant and a tangent at the point of is equal to the angle subtended by the chord in corresponding alternate segment. ∠ AEB = ∠ ABC and ∠ ABD = ∠ AFB Theorem 9 Proof: Using inscribed angle theorem ∠ AEB = m(arc AFB) and using tangent secant theorem ∠ ABC = m(arc AFB)

Converse of theorem 9. Theorem 10

Practice Sums 3

Concyclic Points The opposite angles of a cyclic quadrilateral are supplementary. Theorem 6 Proof: Using inscribed angle theorem, arcs BAD and BCD form complete circle, so their total angle is 360 0

Converse 1.If a pair of opposite angles of a quadrilateral are supplementary, then the quadrilateral is cyclic. Theorem 7

Practice Sums 4

Intersecting Secants AP. BP = CP. PD Theorem 11 Proof: Prove that ∆ APC and ∆ DPB are similar by AA test

Instead of 2 Secants, one is a tangent AP. BP = PT 2 Theorem 12 Proof: Prove that ∆ PTA and ∆ PTB are similar by AA test

Practice Sums 5

Questions and Answers

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