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(1) A circle is the collection of those points in a plane that are at a given constant distance from a fixed-point in the plane. The fixed point is called the centre and the given constant distance is called the radius of the circle.
A Circle with centre O and radius r usually denoted by . Thus, in set theoretic notations, we writeÂ
(2) A point P lies inside or on or outside the circle according as or or r" />.
(3) The collection of all points lying inside and on the circle is called a circular disc with centre and radius .
The set of all points lying inside and on the circle is called a Circular Disc. It is also known as the circular region.
(4) Circles having the same center and different radii are said to be concentric circles.
When two or more circles have the same center but have different radii, they are called as concentric circles, that is, circles with common center.
(5) A continuous piece of a circle is called an arc of the circle.
For Example:Â Consider circle C (O, r). Let be point on the circle. Then, the pieces etc. are all arcs of the circle .
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(6) Prove that If two arcs of circle are congruent, Â then corresponding chords are equal.
Given: Arc PQ of a Circle and arc RS of another circle such that
To Prove:
Construction: Draw Line segment , , and .Proof:
Case-I When Â and are minor Arcs
In triangle and , We have
Â Â Â Â Â Â Â Â Â [Equal radii of two circles]
Â Â Â Â Â Â Â Â Â
So by SAS Criterion of congruence, we have
Case-II When and are major arcs.
If , are major arcs, then and are Minor arcs.
So
Â
Hence,
(7) Prove that If two chords of a circle are equal, then their corresponding arcs are congruent.
Given: Equal chords, of a circle Â and RS of congruent circle
To Prove: , where both and are minor, major or semi-circular arcs.
Construction: If are not diameters, draw line segments , , and .Proof:
Case I: when and are diameters
In this case, and are semi circle of equal radii, hence they are congruent.
Case II: When and are Minor arcs.
In triangles and , we have
and
So by SSS-criterion of congruence, we have
Â Â Â Â Â Â Â Â Â Â Â
Â Â Â Â Â =
Â Â Â Â
Case III: When and are major arcs
In this case, And will be minor arcs.
Â Â Â Â Â Â Â Â Â Â
Â Â Â Â Â =
Â Â Â Â Â
Â Â Â Â Â
Â Â Â Â Â
Hence, in all the three cases, we haveÂ
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(8) Prove that The perpendicular from the centre of a circle to a chord bisects the chord.
Given: A Chord of a circle and perpendicular to the chord .
To Prove:
Construction: Join and Proof: In Triangles and , we have
Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â [Radii of the same circle]
Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â [Common]
And,Â Â Â Â Â Â Â Â Â [Each equal to ]
So, by RHS-criterion of congruence, we have
Â Â Â Â Â Â Â Â Â Â Â
(9) Prove that The line segment joining the centre of a circle to the mid-point of a chord is perpendicular to the chord.
Given: A Chord OF a circle with mid-point M.
To Prove:
Construction: Join and Proof: In triangles and , we have
Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â [Radii of the same circle]
Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â [ is mid-point of ]
So, by SSC- criterion of congruence, we have
Â Â Â Â Â Â Â Â Â Â Â
But , Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â [Linear pair]
Â Â Â Â Â Â Â Â Â Â Â Â []
Â Â Â Â Â Â Â Â Â Â Â
Â Â Â Â Â Â Â Â Â Â Â
(10) Prove that There is one and only circle passing through three given points.
Given: Three non-collinear points and .
To Prove: There is one and only one circle passing through and .
Construction: Join and . Draw perpendicular bisectors and of and respectively. Since and . are not collinear. Therefore, the perpendicular bisectors and are not parallel.
Let And intersect at . Join and .
Proof:Â Since lies on the perpendicular bisector of .
Therefore,
Again, O Lies on the perpendicular bisector of QR.
Therefore,
Thus, Â Â Â Â Â Â (say)
Taking O as the centre draw a circle of radius s. Clearly, passes through P, Q and R. This proves that there is a circle passing the points P,Q and R.
We shall now prove that this is the only circle passing through and .
If possible, let there be another circle with centre and radius r, passing through the points and . Then, will lie on the perpendicular bisectors of and of .
Since two lines cannot intersect at more than one point, so must coincide with O. Since OP=r, and O and coincide, we must have
Â Â Â Â Â Â Â Â Â Â
Hence, there is one and only one circle passing through three non-collinear points and .
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(11) Prove that If two circles intersect in two points, then the through the centre is perpendicular to the common chord.
Given: Two circles and intersecting at points and .
To Prove: is perpendicular bisector of .
Construction: Draw line segments
and Proof: In triangles and ,Â we have
And, Â Â Â Â Â Â Â Â Â Â Â Â Â Â =
So, by SSS-criterion of congruence, we have
Â Â Â Â Â Â
Â Â Â Â Â Â Â Â Â Â Â Â Â Â [ and ]
Let M be the point of intersection of and
In triangles and , we have
Â Â Â Â Â Â
Â Â Â Â Â Â Â Â Â Â Â Â Â [ and ]
Let be the point of intersection of and
In triangles and , we have
And Â Â Â Â Â Â Â Â
So, by SAS-criterion of congruence, we have
Â Â Â Â Â Â and
But, Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â
Â Â Â Â Â Â
Thus,Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â and
Hence, is the perpendicular bisector of .
(12) Prove that Equal chords of a circle subtend equal angle at the centre.
Given: Two Chord AB and CD of circle such that and and
To Prove: Chord AB and CD are equidistant from the centre O i.e .
Construction: Join and .Proof: Since the perpendicular from the centre of a circle to a chord bisects the chord.
Therefore,
..........(i)
And,Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â ..........(ii)
But, Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â
Â Â Â Â Â Â
Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â [Using (i) and (ii) ]â€¦.......(iii)
Now, in right triangles OAL and OCM, we have
Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â [Equal to radius of the circle]
Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â [From equation (iii)]
And, Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â [Each equal to ]
So by RHS criterion of convergence, we have
Â Â Â Â Â Â
Hence, equal chord of a circle are equidistant from the centre.
(13) Prove that Chords of a circle which are equidistant from the centre are equal.
Given: Two Chords AB and CD of a circle which are equidistant from its centre i.e. , where and .
To Prove: Chords are Equal i.e.
Construction: Join OA and OCProof: Since the perpendicular from the centre of a circle to a chord bisects the chord.
Therefore,
Â Â Â Â Â Â
Â Â Â Â Â Â
And, Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â
Â Â Â Â Â Â
Â Â Â Â Â Â
In triangles and , we have
Â Â Â Â Â Â Â Â Â Â Â Â [Each equal to radius of the given Circle]
Â Â Â Â Â Â Â Â [Each equal to ]
And, Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â [Given]
So, by RHS, criterion of convergence, we have
Â Â Â Â Â Â
Â Â Â Â Â Â
Â Â Â Â Â Â
Hence, the chords of a circle which are equidistant from the centre are equal.
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(14) Prove that Equal chords of a circle subtend equal angle at the centre.
Given: A circle and its two equal chords AB and CD.
To Prove: Proof:In triangles AOB and COD, we have
Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â [Given]
Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â [Each equal to r]
Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â [Each equal to r]
So, by SSC-criterion of Congruence, we have
Â Â Â Â Â Â
(15) Prove that If the angles subtended by two chords of a circle at the centre are equal, the chords are equal.
Given: Two Chord AB and CD of a circle C(O,r) such that
To Prove:Â Proof: In triangles AOB and COD, we have
Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â [Each equal to r]
Â Â Â Â Â Â Â Â [Given]
Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â [Each equal to r]
So, by SAS-criterion of congruence, we have
Â Â Â Â Â Â
(16) Prove that Of any two chords of a circle, the larger chord is nearer to the centre.
Given: Two Chord and of a circle with Centre O such that CD" />
To Prove: Chord is nearer to the centre of the circle i.e. , where and are perpendiculars from O to AB and CD respectively
Construction: Join and .Proof: Since the perpendicular from the centre of a circle to a chord bisects the chord. Therefore,
And, Â Â Â Â Â Â Â Â Â Â Â Â Â Â
In right triangles and , we have
And,Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â
Â Â Â Â Â Â .. (i)Â []
Now,Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â CD" />
Â Â Â Â Â Â {1 \over 2}CD" />
Â Â Â Â Â Â CM" />
Â Â Â Â Â Â C{M^2}" />
Â Â Â Â Â Â O{L^2} + C{M^2}" />Â Â Â Â Â Â Â Â [Adding on both sides]
Â Â Â Â Â Â O{L^2} + C{M^2}" />Â Â Â Â Â Â [using equation (i)]
Â Â Â Â Â Â O{L^2} " />
Â Â Â Â Â Â OL" />
Â Â Â Â Â Â
Hence, is nearer to the centre than .
(17) Prove that Of any two chords of a circle,Â the chord nearer to the centre is larger.
Given: Two Chord and of a circle such that , where and are perpendiculars From O on AB and CD respectively.
To Prove: CD" />
Construction: Join and .Proof: Since the perpendicular From the Centre of a circle to a chord bisects the chord.
and
In right triangles OAL and OCM, we have
and,
Â Â Â Â Â Â ....... (i)
And, Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â .......(ii)
Now, Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â
Â Â Â Â Â Â
Â Â Â Â Â Â - O{M^2}" />
Â Â Â Â Â Â O{A^2} - O{M^2}" />Â Â Â Â Â Â Â Â [adding on both sides]
Â Â Â Â Â Â O{C^2} - O{M^2}" />Â Â Â Â Â Â Â Â Â []
Â Â Â Â Â Â C{M^2}" />
Â Â Â Â Â Â CM" />
Â Â Â Â Â Â 2CM" />
Â Â Â Â Â Â CD" />
(18) Prove that The angle subtended by an arc of a circle at the centre is double the angle subtended by it at any point on the remaining part of the circle.
Given: An arc of a circle and a point R on the remaining part of the circle i.e. arc .
To Prove:
Construction: join and produce it to a point M outside the circle.
Proof: We shall consider the following three different cases:
Case I: when is a minor arc.
We know that an exterior angle of a triangle is equal to the sum of the interior oppsite angles.
In , is the exterior angle.
Â Â Â Â [, ] .....................(i)
In , is the exterior angle.
Â Â [, ]
................(ii)
Adding equation (i) and (ii), we get
Â
Case II: when is a semi-circle
We know that an exterior angle of a triangle is equal to the sum of the interior opposite angles.
In , we have
Â Â Â Â [, ]
.....................(iii)
In , We have
Â Â [, ]
................(iv)
Adding equations (iii) and (iv), we get
Case III: When is a major arc.
We know that an exterior angle of a triangle is equal to the sum of the interior opposite angles
In , we have
Â Â Â Â Â Â Â Â Â Â [, ]
...................(v)
In , we have
...................(vi)
Adding equations (v) and (vi), we get
Reflex
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(19)Prove that Angles in the same segment of a circle are equal.
Given: A circle , an arc PQ and two angles and in the same segment ofÂ the circle.
To Prove: =
Construction: Join and Proof: we know that the angle subtended by an arc at the centre is double the angle subtended by the arc at any point in the remaining part of the circle. So we have
and
We have
Reflex and
Thus , in both the cases, we have
(20) Prove that The angle in a semi-circle is a right angle.
Given: is a diameter of a circle and is an angle in semi-circle.
To Prove: Proof: we know that the angle subtended by an arc of a circle at its centre is twice the angle formed by the same arc at a point on the circle. So, we have
Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â [ is a straight line]
(21) Prove that The opposite angles of a cyclic quadrilateral are supplementary.
Given: A Cyclic quadrilateral ABCD
To Prove: and
Construction: Join and .
Proof: Consider side of quadrilateral ABCD as the Chord of the circle. Clearly, and are angles in the same segment determined by chord of the Circle.
Â Â Â Â Â Â Â Â Â â€¦â€¦â€¦â€¦.(i)
Now , consider the side of quadrilateral ABCD as the chord of the circle. We find that Â and are angles in the same segment
Â = Â Â Â Â Â Â Â Â Â Â Â Â [angles in the same segment are equal]..(ii)
Adding equation (i) and (ii), we get
Â Â Â Â Â Â
Â Â Â Â Â Â
Â Â Â Â Â Â
Â Â Â Â Â Â Â Â Â Â Â Â Â Â [sum of angle of triangle is ]
Â Â Â Â Â Â
Â Â Â Â Â Â
But,
Hence, and
The converse of this theorem is also true as given below.
(22) Prove that If the sum of any pair of opposite angles of a quadrilateral is ,then it is cyclic.
Given: A quadrilateral ABCD in which
To Prove: ABCD is acyclic quadrilateral.Proof: If possible, Let ABCD be not cyclic quadrilateral. Draw a circle passing through three non-collinear points , and . Suppose the circle meets or produced at . Join .
Now, ABCDâ€™ is a cyclic quadrilateral.
..............(i)
But,
i.e. ..............(ii)
from (i) and (ii), we get
=
Â Â Â Â Â Â =
Â Â Â Â Â Â An exterior angle of is equal to interior oppsite angle.
But, this is not possible, unless coincides with . Thus, the circle passing through also passes through .
Hence, ABCD is a cyclic Quadrilateral.
(23) Prove that If one side of a cyclic quadrilateral is produced,Â then the exterior angle is equal to the interior opposite angle.
Given: A Cyclic quadrilateral one of whose side is produced to .
To Prove: Proof: Since is a quadrilateral and the sum of opposite pairs of angles in a cyclic quadrilateral is
But, Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â [Liner Pairs]
Â Â Â Â Â Â
Or, Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â
(24) Prove that The quadrilateral formed by angle bisectors of a cyclic quadrilateral is also cyclic.
Given: A Cyclic quadrilateral in which and are the bisectors of , , and respectively such that a quadrilateral is formed.
To Prove: is a cyclic quadrilateral.Proof: In order to prove that is a cyclic quadrilateral, it is sufficinet to show that
Since the sum of the angles of a triangle is . Therefore, in triangles and , we have
And, Â Â Â Â Â Â Â Â Â Â Â Â Â Â
Â Â Â Â Â Â
And,Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â
Â Â Â Â Â Â
Hence, is a cyclic Quadrilateral.
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(25) Prove that If two sides cyclic quadrilateral are parallel, then the remaining two sides are equal and the diagonals are also equal.
Given: A Cyclic quadrilateral in which .
To Prove: (i) Â Â Â Â Â Â Â Â Â (ii) Proof: In order to prove the desired results, it is sufficient to show that . Since is cyclic Quadrilateral and sum of oppsite pairs of angles in a cyclic Quadrilateral is
........(i)
Since and BC is a transveral and sum of the interior angles on the same side of a transversal is
...................(ii)
From (i) and (ii), we get
Â Â Â Â Â Â .................(iii)
Now, consider triangles ADC and BCD. In and , we have
Â Â Â Â Â Â Â Â Â Â [From equation (iii)]
Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â [Common]
And, Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â [ and are angles in the segment of chord CD]
So, by AAS-criterion of congruence, we have
Â Â Â Â Â and
(26) Prove that If two opposite sides of a cyclic quadrilateral are equal, then the other two sides are parallel.
Given: A cyclic quadrilateral such that .
To Prove:
Construction: Join .Proof: We have,
Â Â Â Â Â Â
Â Â Â Â Â Â
Â Â Â Â Â Â
Â Â Â Â Â Â
But, these are alternate interior angles. Therefore, .
(27) Prove that An isosceles trapezium is cyclic.
Given: A trapezium in which and
To Prove: is a cyclic trapezium.
Construction: Draw and .Proof: In order to prove that ABCD is a cyclic trapezium, it is sufficient to show that .
In triangles and , we have
Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â [Given]
Â Â Â Â Â Â Â Â Â Â [Each equal to ]
And, Â Â Â Â Â Â Â Â Â Â Â Â Â Â
So, by RHS-criterion of congruence, we have
Â Â Â Â Â Â and
Now, Â Â Â Â Â Â Â Â Â Â Â Â Â
Â Â Â Â Â Â
Â Â Â Â Â Â
Â Â Â Â Â Â
Â Â Â Â Â Â
Thus, Â Â Â Â Â Â Â Â Â Â Â Â Â and .
Â Â Â Â Â Â
Â Â Â Â Â Â
Hence, ABCD is a cyclic quadrilateral.
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