Page 23 - 12 Math 6

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∴ OM=2  y1 + y2 - 0 2 +  x1 + x2 - 0=2 y12 + y22 + x12 + x22 + 2x1x2 + 2 y1 y2
 2   2 4

= (x12 + y12 ) + (x22 + y22 ) + 2x1x2 + 2 y1 y2
4
= a2 + a2 + 2x1x2 + 2 y1y2
4 ( A and B lie on the circle.)

OM 2 = 2a2 + 2x1x2 + 2 y1y2
4

= a2 + x1x2 + y1y2 (1)
2

Similarly ON 2 = a2 + x3x4 + y3 y4 ( chords (2) congruent)
are
2

We know that AB 2 = CD 2
or (x2 - x1)2 + ( y2 - y1)2 = (x4 - x3 )2 + ( y4 - y3)2

or x22 + x12 + y22 + y12 - 2x1x2 - 2 y1 y2 = x42 + x32 - 2x3x4 + y42 + y32 - 2 y3 y4

or a2 + a2 - 2x1x2 - 2 y1 y2 = a2 + a2 - 2x3x4 - 2 y3 y4 ( x12 + y12 = a2 etc)

or 2a2 - 2x1x2 - 2 y1y2 =2a2 - 2x3x4 - 2 y3 y4 Challenge!

or x1x2 + y1 y2 = x3x4 + y3 y4 (3) State and prove the

or OM 2 = ON 2 converse of this Theorem.

Theorem 6: Show that measure of the central angle of a
minor arc is double the measure of the angle subtended in the
corresponding major arc.

Proof: Let the circle be x2 + y2 = a2.
A(a cosq1 , a sinq1) and B(a cosq2 , a sinq2) be end points of a
minor arc AB. Let P (a cosq , a sinq) be a point on the major arc.

Central angle subtended by the minor arc AB is ∠AOB = q2 - q1.

We need to show m∠APB = 1 (q2 - q1)
2

version: 1.1

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