Before you start learning about circles related to 9th grade curriculum, please revise what we learnt in 8th grade. 8th Grade Revision – Circles.
Congruence of Circle and Arcs
Congruent Circles:
Two circles are said to be congruent if and only if either if them can be super imposed on the other so as to cover it exactly.
This means that two circles are congruent if and only if their radii are equal. i.e. circles and
are congruent if
.
Congruent Arcs:
Two arcs of a circle are congruent if either of them can be super imposed on the other so as to cover it exactly.
This happens if degree measures of two arcs are the same. Therefore we can say that two arcs of a circle (or of congruent circles) are said to be congruent if and only if they have the same degree measure.
If two arcs and
are convergent arcs if a circle
, we write
.
Thus,
Important results and their proofs on congruent arcs and chords:
Theorem 1: If two arcs of a circle (or of congruent circles) are congruent, then the corresponding chords are equal.
Explanation: Follow the diagram shown:
Given: Two circles and
. They are congruent hence their radius is equal.
Also .
Construction: Join and
to
in the first circle and
and
to
in the second circle.
To Prove: (i.e. corresponding chords are equal).
There can be two cases: When and
are minor arcs (case 1) and When
and
are major arcs (case 2).
Case 1: and
are minor arcs
Consider and
(radius of the two circles are equal)
Also
(Since )
Hence (by SAS theorem)
. Hence proved.
Case 2: and
are major arcs
From case 1, we proved that that
Hence
Theorem 2: If two chords of a circle (or of congruent circles) are equal, then the corresponding arcs (major, minor or semi-circular) are congruent.
Explanation: Follow the diagram shown:
Given: Chords of circle
and
of circle
are equal.
Construction: Join and
to
in the first circle and
and
to
in the second circle.
and
are not diameters.
To Prove: where they can be both minor, major or semi-circular arcs.
Proof: There are three possible cases.
Case 1: When and
are diameters.
In this case, and
are semi circles of equal radii. Hence they are congruent.
Case 2: When and
are minor arcs.
Consider and
Given:
Therefore (By S.S.S criterion)
Case 3: When and
are major arcs
and
are minor arcs
Therefore in all three cases
Theorem 3: The perpendicular from the center of a circle to a chord bisects the chord.
Given: is a chord in Circle
and
To Prove:
Construction: Join and
Proof: Consider and
(given)
is common
Therefore (By S.S.A criterion)
Therefore
Hence bisects
Theorem 4 (converse of Theorem 3): If the line joining the center of the circle to the mid point of the chord is perpendicular to the chord.
Given: is the midpoint of
is a chord in circle
To Prove:
Proof: Consider and
is common
(given)
Therefore (By S.S.S criterion)
Therefore
Since (linear pairs)
Theorem 4.1: The perpendicular bisectors of two chords of a circle intersect at its center.
Given: and
are chords in circle
(
is the midpoint of
)
(
is midpoint of
)
and
Let and
meet at
To Prove: coincides with
Construction: Join and
Proof: Since is the mid point of
(Theorem 4)
is perpendicular bisector of
and
are perpendicular bisector of
lies on
Similarly, is the mid point of
is perpendicular bisector of
(Theorem 4)
and
are perpendicular bisector of
lies on
Therefore, we can say (point of intersection of
and
) coincides with
(center of the circle)
Therefore, Perpendicular bisector of and
will intersect at the center of the circle.
Theorem 5: There is one and only one circle passing through three non-collinear points.
Given: and
are three non-collinear points.
To Prove: One and only one circle passes through and
Construction: Join and
Draw perpendicular bisector of
and also draw perpendicular bisector
of
Let and
intersect at
Proof: Since lies on perpendicular bisector of
Similarly,
(radius of the circle)
Now taking as the center and
as the radius, draw a circle.
You will see that the circle passes through the points and
.
We now need to prove that this is the only circle passing through the points and
.
Now let us assume that there is another circle with center latex P, Q $ and
.
Then will lie on the perpendicular bisector of
of
and
of
Since two lines cannot intersect in two points, must coincides with
Hence only one and one circle can pass through three non-collinear points.
Note 1: In a case where the three points are collinear, then a single circle cannot pass through all of them.
Note 2: This theorem also implies that there will be a unique circle that passes through the vertices of a triangle. This circle is called circum-circle of the triangle and the center of the circle is called circum-center.
Note 3: Infinite number of circles can be drawn through one point.
Note 4: Infinite number of circles can be drawn through two points.
Theorem 6: If two chords AB and AC of a circle with center O are equal then the center of the circle lies on the angle bisector of .
Given: are chords of circle
To Prove: lies on the angle bisector of
Construction: Join . Draw bisector of
Proof: Consider and
(given)
is common
(By construction)
Therefore (By S.A.S criterion)
and
Since
is perpendicular to
and it bisects
passes through the center
(By theorem 4)
Theorem 7 (converse of Theorem 6): If two chords and
of a circle with center
are such that the center
lies on the bisector of
, then
i.e. chords are equal.
Given: and
are chords of circle
To Prove:
Construction: Join and
Proof: In
(radius)
(angles opposite to equal sides are equal)
Similarly, in
(radius)
(angles opposite to equal sides are equal)
But
Consider and
is common
Therefore
(since corresponding parts of congruent triangles are equal)
Theorem 8: If two circles intersect in two points, then the line through the centers of the circles is perpendicular bisector of the common chord.
Given: Two circles and
They intersect at and
Construction: Join and
. Also join
and
To Prove: is perpendicular bisector of
Proof: Consider and
is common
Therefore (By S.S.S criterion)
Let and
intersect at
Consider and
is common
Therefore (By S.A.S criterion)
and
We know
Thus and
Hence is perpendicular bisector of