Flash Player 9 (or above) is needed to view presentations.
We have detected that you do not have it on your computer. To install it, go here.

Like this document? Why not share!

# Lesson 30 35

## on Dec 01, 2010

• 1,563 views

### Views

Total Views
1,563
Views on SlideShare
1,534
Embed Views
29

Likes
1
34
0

### 2 Embeds29

 http://scallions11precalc.blogspot.com 28 http://www.scallions11precalc.blogspot.com 1

### Report content

• Comment goes here.
Are you sure you want to

## Lesson 30 35Document Transcript

• Lessons 30 -35Information on Circles
• Lesson 30 & 31 Notes Terminology Circle with centre k diameter semicirclesk chord Radius from centre to outside edge Perpendicular bisector central angle Inscribed angle tangent point of tangency The properties that should be discovered and used are as follows:
• A) The perpendicular (line) from the centre of a circle to a chord bisects the chord This also works in the reverse. If you find the middle point of any chord and draw a perpendicular line from the middle point it will pass through the center of the circle. m AEC = 90.00 A C EC = 3.19 cm FE = 3.19 cm E FThis property can help you find the centerbecause, if you apply this property more thanonce inside one circle, the point where the Centerperpendicular lines intersect is where thecenter of the circle is. FB) The measure of the central angle is equal to twice the measure of the inscribed anglesubtended (shared) by the same arc.This property becomes very useful forfinding the center when we start to This anglecreate central angles of 1800. This means will bethat the inscribed angle will be 900. half the central This Angle will be twice as much as the inscribed The Arc that is shared is called the Subtended Arc
• If students create a 900 angle on the circumference of the circle the end of each raywill be the diameter. If you apply this property more than once, the point wherethe diameters insect will be the center of thecircle.C) The inscribed angles subtended by the same arc are congruent. Again, this is a property that only works M K when the inscribed angles share an arc. m JML = 42.94 If you are using computer software, ask m JKL = 42.94 the students to drag one of the vertices into the subtended arc and see what happens to the measurements. They will no longer be congruent because they are no longer sharing the same arc. L JD) A tangent to a circle is perpendicular to the radius at the point oftangency. This becomes a useful tool for finding the center of a circle. If we extend the radius to a diameter, and apply the property more than once, the intersection point will be the center.
• E) An angle inscribed in a semi-circleis ALWAYS 900.This is directly related item B in thelist. Do this twice and the point wherethe hypotenuses intersect will be thecenter of the circle.F) The opposite angle of a cyclicquadrilateral (are quadrilaterals inside a circle with its vertices on thecircumference of the circle) are supplementary (add up to 1800). H I G 180 I G JG) An exterior angle of a cyclic quadrilateral is equal to the interior oppositeangle. HJG 40 and HGK 110 H I Then we know that angle JGH = 700 because it is in a straight line K with HGK (180 – 110 = 70) From this we can then figure out G that GHJ 70 because interior angles of GHJ 180 J If we know that GHJ 70 we also know that GIJ 70 because both are inscribed angles subtended by the minor arc JG.
• Examples ECB BDC 1) Determine the measures of angles where E is the center of the circle. BAD DBE A B D E 100 30 C 2) Determine the length of a chord AC. B 10 6 C A 3) Given a circle with the center O, and OF CD and DE=20 and OF=DF find the length of OF EO F DFC D CD OE
• Lesson 32 – Properties of Tangents1) Tangent segments to a circle’s circumference from anyexternal point are congruent in length.2) The angle between the tangent and the chord ishalf the measure of the intercepted arc.3) The angle between the tangent and the chord is equal to the inscribed angle onthe opposite side of the chord.Examples1) The circle has a radius of 5 and AB is tangent to the circle at point C. B 50 and OA = OB O Find the Area of AOB A C B
• A2) 50 Find the length of AC 5 30 D C B3) Given that ED is tangent at c BCA 60 ACD (2 x) BCE (3x 5) Find B 60
• AD BC4) ACB 15 Find DBCLesson 33 – No Notes – Just ReviewLesson 34 – Circle and Polygon PropertiesExample 1 – If diameter CD is perpendicular to a chord AB at E, verify that ABC is isosceles.This means we want to prove AC=CB
• Example 2)Determine the measure of BAC if DEF 60 and EFC 70Example 3)Find the center of the circle that passes through the points (0,0), (0,6) and (4,0). 8 6 4 2 -15 -10 -5 5 10 15 -2 -4 -6 -8 -10
• Example 4)With a circle centered at (0,0) and tangent to the circle at (3,4) find the equation ofAB 10 8 6 4 2 -15 -10 -5 5 10 15 -2 -4 -6 -8