e math reporting

8/18/2019 E Math Reporting

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Circles Determined byGeometric ConditionsPresented by: Group 4 (9G E-Math)


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Learning Objectives:

To determine what are the geometricconditions determining Circes

To be abe to answer !uestions in rea

to the topics to be discussed To understand the de"nitions and its

to the conditions gi$en on Circes


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Geometric Conditions

Circe Tangent to a %ine

&ntersection between the %ine

the Circe Points o# &ntersection o# Two C


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Word VocabularyCirce ' et o# points on the pane that are e!uidistant #rom a "ed point c

Tangent %ine - + ine that touches a circe at ony one point*

- &t is a ine perpendicuar to the radius o# a circe*

,adius - + ine segment that determines the distance #rom the center tothe circe* - &ts pura #orm is ,adii (ray-dee-eye)

ecant ' + ine intersecting a circe at two points*


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Word Vocabulary

Center ' The point around which a circe is desc

Cartesian Coordinate ystem - a coordinate systspeci"es each point uni!uey in a pane by a panumerica coordinates*


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CIRCLE T!GE!T TO LI!EC&,C%E ETE,M&.E /0 GE1MET,&C C1.&T&1.


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CIRCLE T!GE!T TO LI

Eampe 23

Find the equation of a circle if the circle is tangent to the

line –x + y + 4 = 0 at the point (3, -1), and the center is on th

line x + 2y – 3 = 0 .


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oution:

Let C(x, y) and radius = r. Line

perpendicular to the line –x + y + 4 = 0

through (3, 1) passes through the

center of the circle

Coordinates of the center is the point of

intersection between x + 2y – 3 = 0 and

x + y – 2 = 0.

Solve for x and y,


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T"o met#ods to $nd tradius:

Method 1:Using the directed distance between –x + y + 4 = 0

point (1, 1)

Method 2:Using the distance formula between (3, 1) and (1,


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oution:

Therefore,

Simplifying,


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CIRCLE T!GE!T TO LI

Eampe 2


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oution:rag picture to pacehoicon to add


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I!TER%ECTIO!

&ETWEE! LI!E !D CIRCLEC&,C%E ETE,M&.E /0 GE1MET,&C C1.&T&1.


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I!TER%ECTIO! &ETWEE! LI!E !D

Eampe 23

Find the points of intersection between the line x =

the circle x 2 + y 2 + 2x – 2y = 2 . Draw the figure


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oution:

Substitute the equation of the line to

the equation of the circle.

Therefore the points of

intersections are ,


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I!TER%ECTIO! &ETWEE! LI!E !D

Eampe 2


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oution:rag picture to pacehoicon to add


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'OI!T% O(

I!TER%ECTIO! O(TWO CIRCLE%C&,C%E ETE,M&.E /0 GE1MET,&C C1.&T&1.


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'OI!T% O( I!TER%ECTIO! O( TWO C

Eampe 23

ind the points of intersection of the circles x 2 + y 2 – 2x

and x 2 + y 2 + 2x + 2y – 2 = 0 ! "raw the circles!


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oution:

Subtracting,

Substituting,

Find y if x = 1 and x = -1,

Therefore, the points of intersection

are


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'OI!T% O( I!TER%ECTIO! O( TWO C

Eampe 2

#i$en the e%uations, x 2 + (y - 2) = 10 and (x – 2) + y 2 = 1

points of intersection of the circles


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oution:

x2 + y2 – 4y + 4 = x2 – 4x + 4 + y2

Combine all lie terms,

4x – 4y = 0

!aving an answer of,

x = y

Find y if x " y

y = 3, -1Therefore, the points of intersection

are,

P1 (3, 3) P2 (-1, -1)


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T6+.7 018

1,

%&TE.&.G

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