## Circle Theorem Maths Notes

**Basic Concepts of Circle:**

Definition: The set of points equidistant from a fixed point in a plane is a circle.

The fixed point is called the centre of the circle. The segment joining the centre of the circle and a point on the circle is called the radius of the circle.

In the figure, 0 is the centre of the circle and seg OP is the radius. Circle can have infinite radii. Radii of the same circle are congruent.

The segment joining any two points of the circle is called a chord. In the figure seg AB is the chord. Circle can have infinite chords.

The chord which passes through the centre of a circle is the diameter. Diameter is the longest chord of the circle. Diameter is twice the radius. A circle can have infinite diameters.

**A perpendicular drawn from the centre of a circle on its chord bisects the chord.**

In the figure, if O is the centre of the circle, seg 0M ⊥ chord AB and A-M-B; then AM = MB.

**The segment joining the centre of a circle and the midpoint of Its chord is perpendicular to the chord.**

In the figure, if O is the centre of the circle, M is the midpoint of chord PQ, then seg OM ⊥ chord PQ.

i.e. ∠OMP = ∠OMQ = 90°.

**Congruent chords of a circle are equidistant from the centre of the circle.**

In the figure, O is the centre of the circle. If chord AB ≅ chord CD, seg OM ⊥ chord AB and seg ON ⊥ chord CD such that A-M-B and C-N-D then 0M = ON.

**The chords of a circle equidistant from the centre of a circle are congruent.**

In the figure, O is the centre of the circle. seg PQ and seg RS are chords.

seg OM ⊥ chord RS and seg ON ⊥ chord PQ such that R-M-S and P-N-Q.

If OM = ON, then chord PQ ≅ chord RS.

**Read and Understand:**

- Infinite number of circles can pass through one point.
- Infinite circles pass through two distinct points.
- There is one and only one circle (i.e. unique circle) passing through three non-collinear points.
- No circle can pass through three coimear points.

**Secant and Tangent:**

In the figure, line m and the circle have one and only one point P in common. Such a line is called the tangent of the circle and the common point is called the point of contact. In the figure, point P is the point of contact.

In the figure, line l and the circle have two points in common. Such a line is called as a secant of the circle. Points A and B are the points of intersection of the circle and the secant.

In the figure, if we draw the radius OP, it will be perpendicular to line m.

This concept is known as Tangent theorem.

**Tangent theorem:**

Statement: A tangent at any point of a circle is perpendicular to the radius at the point of contact.

Given:

(1) A circle with centre O

(2) Line l is tangent to the circle at point P.

To Prove: Line ⊥ radius OP

Proof: (Indirect method)

Let us assume thal line l is not perpendicular to seg OP then let seg OQ⊥ line l and let point Q lie on line l. P and Q are different points.

Take a point R on the line l such that R – Q – P and QR = QP. Draw seg OR.

∴ OR = OP

seg OP is the radius

∴ seg OR is the radius.

∵ point O is the centre,

∴ points P and R lie on the circle.

But they also lie on line l.

∴ line l which is tangent has two points P and R in common with the circle.

This contradicts the definition of tangent.

∴ our assumption is false.

∴ line l ⊥ radius OP.

**Converse of tangent theorem:**

Statement: A line perpendicular to a radius at its point on the circle is a tangent to the circle.

Given:

- A circle with centre 0
- seg OP is the radius
- Line l ⊥ radius OP at point P.

To prove: line l is a tangent to the circle.

Proof: Take a point Q on line l, other than P.

Draw seg OQ.

In ∆ OPQ, ∠ OPQ = 90°.

∴ ∆ OPQ is a right angled triangle.

In a right angled triangle, the hypotenuse is the largest side.

∴ OQ > OP

∴ OQ > radius … (∵ OP = radius)

∴ Q doesn’t lie on the circle.

This is true for all the points on line l, other than P.

∴ only point P is the common point between line l and the circle.

∴ line I is tangent to the circle … (By definition)

Note:

- A circle can have infinite number of tangents.
- One and only one tangent can be drawn at a given point on the circle.
- No tangent can be drawn from a point in the interior of the circle to that circle.
- Two tangents can be drawn to a circle from the point in the exterior of the circle.

**Read and Understand:**

In the figure, two tangents PQ and PR are drawn from an external point P to the circle with centre O. Seg PQ and seg PR are called as tangent segments.

**Tangent Segment Theorem:**

Statement: Tangent segments drawn from an external point to a circle are congruent.

Given:

- A circle with centre O.
- Lines PQ and PR are tangents to the circle at points Q and R respectively.

To prove: seg PQ ≅ seg PR

Construction: Draw seg OP, seg OQ and seg OR

Proof: In ∆ OQP and ∆ ORP,

**Remember this!**

- Tangent theorem: The tangent at any point of a circle is perpendicular to the radius through the point of contact.
- Converse of tangent theorem: A line perpendicular to a radius at its point on the circle is tangent to the circle.
- Tangent segment theorem: Tangent segments drawn from an external point to a circle are congruent.

**Touching circles:**

If two circles in the same plane have only one point in common are called touching circles. Circles can touch each other internally or circles can touch each other externally.

In figure 1, the circles with centres O and P touch each other externally at point T. Line l is the common tangent at point T.

In figure 2, the circles with centre O and P touch each other internally at point T. Line l is the common tangent at point T.

**Remember this!**

When circles touch each other externally, the centres lie on the opposite side of the common tangent. Whereas, when circles touch each other internally, the centres lie on the same side of the common tangent.

**Theorem of touching circles:**

Statement: If two circles touch each other, their point of contact lies on the line joining the centres.

Given: Two circles with centres O and P touch each other at point T.

To prove: Point T lies on line OP.

Construction: Draw seg OT, seg PT and common tangent l at point T.

Proof: seg OT ⊥ line l … (Tangent theorem) … (1)

seg PT ⊥ line l … (Tangent theorem) … (2)

∴ seg OT and seg PT are perpendiculars to line l at point T.

∵ one and only one perpendicular ‘can be drawn at a given point on a line.

∴ lines OT and PT are not two different lines.

They are one and the same.

∴ points O, T and P are collinear.

∴ point T lies on line OP.

**Remember this!**

- The point of contact of the touching circles lies on the line joining their centres.
- If two circles touch each other externally, distance between their centres is equal to the sum of their radii.
- If two circles touch each other internally, distance between their centres is equal to the difference of their radii.

**Read and Understand:**

**Arc of a circle:**

A secant divides a circle in two parts. Each part together with the points common to the circle, is called an arc. An arc is a part of a circle.

In the figure 1, secant l intersects the circle at points A and B. Arc AXB and arc AYB are thus two arcs formed due to secant 1.

If the secant passes through the centre, each arc so formed is a semicircle.

In the figure 2, line l passes through centre O intersecting the circle at points P and Q. Seg PQ is the diameter. Arc PRQ and arc PSQ are semicircles.

in the figure 1, line l doesn’t pass through the centre. The arc on the side of the centre is the major arc and arc on the other side of the centre is the minor arc. Arc AYB is the major arc and arc AXB is the minor arc.

**Naming an arc:**

Usually three points are used to name an arc. The first and the last points in naming an arc are always its endpoints. In the figure 1, arc AXB and arc AYB have endpoints A and B. The minor arc is named AXB and major arc is named AYB. In case, if an arc is denoted with two points then it is always the minor arc.

**Central Angle:**

An angle in the plane of the circle with its vertex at the centre is called central angle.

In the figure, 0 is the centre of the circle and ∠POQ is the central angle.

Like secant, a central angle also divides a circle into two arcs.

**Measure of an arc:**

m (arc AXB) denotes the measure of arc AXB.

1. Measure of a manor arc is equal to the measure of its corresponding central angle.

In the figure, 0 is the centre of the circle and arc APB is minor arc. Arc AQB is major arc.

Here m(arc APB) = ∠AOB = θ.

2. Measure of major arc

= 360° – measure of corresponding minor arc Here m (arc AQB) = 360° – m (arc APB).

3. Measure of a semicircular arc, that is of a semicircle is 180°.

4. Measure of a circle is 360°.

**Congruence of arcs:**

Two arcs are congruent, if their measures and radii are equal.

OR

Two arcs of the same circle or of congruent circles, having equal measures, are congruent.

If arc PQ and arc EF are congruent then symbolically it is written as arc PQ arc EF.

**Property of sum of measures of arcs:**

In the figure, points P, Q, R, S and T lie on the same circle. Arc PQR and arc RST have only one point R in common.

∴ m(arc PQR) + m(arc RST) – m(arc PRT)

Now, arc PQR and arc QRS have many points in common (i.e. all the points on arc QR).

∴ m(arc PRS) ≠ m(arc PQR) + m(arc QRS).

In figure, chord AB ≅ chord CD, Prove that, arc AC ≅ arc BD.

Proof:

Chord AB ≅ chord CD … (Given)

∴ Arc ACB ≅ arc CBD …… (Arcs corresponding to congruent chords)

∴ m(arc ACB) = m(arc CBD) …….. (1)

But m(arc ACB) = m(arc AC) + m(arc CB) … (2)

and m(arc CBD) = m(arc CB) + m(arc BD) … (3)

From (1), (2) and (3), we get,

m(arc AC) + m(arc CB) m(arc CB) + m(arc BD)

∴ m(arc AC) = m(arc BD)

∴ arc AC ≅ arc BD.

**Inscribed angle:**

In the figure, 0 is the centre of the circle. Point Q lies on the circle. The arms of ∠ PQR intersect the circle at, points P and R. Such an angle is called an Inscribed angle.

In the figure ∠PQR is inscribed in arc PQR.

**Intercepted arc:**

Observe the following figures:

In each of these figures, the arc of the circle which lies in the interior of ∠ PQR is an arc intercepted by ∠ PQR. Both the sides of the angle should contain an endpoint of the arc.

In the figures (a), (b) and (c) ∠ PQR intercepts only one arc whereas in figures (d), (e) and (f) ∠ PQR intercepts two arcs.

Observe figure (g), arm QP does not contain any endpoint of the arc, hence arc is not intercepted by ∠ PQR in this case.

**Inscribed angle theorem:**

Statement : The measure of an inscribed angle is half of the measure of the arc intercepted by It.

Given : In a circle with centre O, ∠ BAC is inscribe in an arc BAC. ∠ BAC intercepts arc BXC of the circle.

To prove : m∠ BAC = \(\frac{1}{2}\) m (arc BXC).

Proof : There arise three cases as shown in parts (a), (b), (c) of the f’igure.

Case 1: The centre is on the angle [as in figure (a)]

In this case, join C to O.

∆ OAC is isosceles with OA = OC.

Let m∠OAC = m∠OCA = p,

By remote interior angles theorem,

m∠BOC = m∠OAC + m∠OCA

m∠BOC = p + p = 2p = 2m∠BAC.

∴ m∠ BAC = \(\frac{1}{2}\) m∠ BOC

But m ∠ BOC = m (arc BXC) … (Definition of measure of minor arc)

∴ m∠BAC = \(\frac{1}{2}\) m(arc BXC).

Case 2: The centre is in the interior of the angle [as in figure (b)]. In this case let D be the other endpoint of the diameter drawn through A.

Let arc CMD be the one intercepted by ∠ CAD, and let arc BND be the one intercepted by ∠ DAB. Then as proved in the first situation, we have

m∠CAD = \(\frac{1}{2}\) m(arc CMD)

and m∠DAB = \(\frac{1}{2}\) m(arc BND)

∴ m∠C + m∠DAB = \(\frac{1}{2}\) m(arcCMD) + \(\frac{1}{2}\) m(arcBND)

∴ m∠BAC = \(\frac{1}{2}\) [m(arc CMD) + m(arc BND)]

= \(\frac{1}{2}\) m(arc BXC).

Case 3: The centre is in the exterior of the angle [as in figure (c)]. Again let D be the other endpoint of the diameter drawn through A. Let arc CMD be the one intercepted by ∠ CAD and let arc BND be the one intercepted by ∠ DAB.

Then as proved in the first situation, we have,

m∠CAD = \(\frac{1}{2}\) m(arc.CMD)

and m∠DAB = \(\frac{1}{2}\) m(arc BND)

∴ m∠CAD – m∠DAB = \(\frac{1}{2}\) m(arc CMD) – \(\frac{1}{2}\)m(arc BND)

∴ m∠BAC = \(\frac{1}{2}\) [m(arc CMD) – m(arc BND)]

= \(\frac{1}{2}\) m(arc BXC).

**Corollaries of inscribed angle theorem:**

Statement: Angles inscribed in the same arc are congruent

Given:

- A circle with centre O
- ∠ABD and ∠ACD are inscribed in arc ABD and intercepts arc APD.

Statement: Angle inscribed in a semicircle is a right angle.

Given:

- A circle with centre O.
- seg AC is the diameter
- ∠ABC is inscribed in arc ABC and intercepts arc AMC.

To prove: ∠ ABC = 90°

Proof:

m(arc AMC) = 180° … (Measure of a semicircle is 180°) (1)

∠ ABC = \(\frac{1}{2}\) m (arc AMC) … (Inscribed angle theorem)

∴ ∠ABC = \(\frac{1}{2}\) × 180° … [From(1)]

∴ ∠ABC = 90°.

**Cyclic quadrilateral:**

A quadrilateral whose all vertices lie on the same circle is a cyclic quadrilateral.

In the figure, points A, B, C and D lie on the same circle.

∴ ▢ABCD is cyclic.

**Theorem of cyclic quadrilateral:**

Statement: Opposite angles of a cyclic quadrilateral are supplementary.

Given: ▢ABCD is cyclic.

To prove: ∠DAB + ∠DCB = 180°

∠ ABC + ∠ ADC = 180°

Proof: ∠DAB is inscribed in arc DAB and intercepts arc DCB.

**Corollary of cyclic quadrilateral theorem:**

Statement: An exterior angle of a cyclic quadrilateral is congruent to the angle opposite to its adjacent interior angle.

Given: ▢ABCD is cyclic. ∠DCE is an exterior angle of ▢ABCD.

To prove: ∠ DCE ≅ ∠ BAD

Proof:

∠DCE + ∠BCD = 180° … (Linear pair of angles) … (1)

▢ABCD is cyclic.

∴ ∠BAD + ∠BCD = 180° … (Theorem of cyclic quadrilateral) … (2)

∴ from (1) and (2), we get, .

∠ DCE + ∠BCD = ∠BAD + ∠BCD

∴ ∠ DCE = ∠BAD

∴∠ DCE ≅∠BAD.

**Converse of cyclic quadrilateral theorem:**

Statement: If a pair of opposite angles of a quadrilateral is supplementary, the quadrilateral is cyclic.

Given: In ▢ABCD, ∠DAB + ∠BCD = 180°.

To prove: ▢ABCD is cyclic.

Proof: There is one and only one circle that passes through three non-collinear points.

∴ A circle passes through D, A and B. Let us assume that the fourth vertex C does not lie on the circle.

∴ it lies either in the interior or in the exterior of the circle.

Suppose C lies in the exterior of the circle, let E be the point where !ine DC intersects the circle. Draw seg BE.

This contradicts statement (3).

∴ point C cannot lie in the exterior of the circle.

Similarly, it can be shown an assumption of C in the interior of the circle also leads to contradiction.

∴ C cannot lie in the interior of the circle

∴ point C has to lie on the circle.

∴ points A, B, C and D lie on the same circle.

∴ ▢ABCD is cyclic.

**Theorem:**

If two points on a given line subtend equal angles at two distinct points which lie on the same side of the line, then the four points are concyclic.

Given: Points S and R lie on the same side of line PQ such that ∠PSQ ≅ ∠PRQ.

To prove: Points P, S, R, Q are concyclic.

Proof: There is one and only one circle that passes through three non-collinear points.

∴ a circle passes through points P, S and Q. Let us assume that thé fourth point R does not lie on the circle.

∴ it lies either in the interior or in the exterior of the circle.

Let us assume that point R lies in the exterior of the circle. Let seg PR intersect the circle at point T. Draw seg QT.

This contradicts (3)

∴ our assumption is false.

∴ point R cannot lie outside the circle.

Let us assume that point R lies in the interior of the circle.

Produce PR to intersect the circle in the point T. Draw QR.

∠PSQ ≅∠PTQ … (Angles inscribed in the same arc are congruent)

But ∠PSQ ≅ ∠PRQ … (Given) … (5)

∴ from (4) and (5),

∠PTQ ≅∠PRQ

∠PRQ is an exterior angle of ∆RTQ,

∠PRQ > ∠ RTQ … (Exterior angle theorem)

i.e. ∠PRQ > ∠PTQ … (P-R-T)

This contradicts (6)

∴ point R cannot lie inside the circle.

∴ point R lies on the circle.

∴ points P, S, R and Q are concyclic.

[Note : The above theorem is the converse of ‘Angles inscribed in the same arc are congruent .]

**Remember this!**

1. Inscribed Angle theorem : The measure of an inscribed angle is half the measure of the arc intercepted by it.

Corollary 1 : Angles inscribed in the same arc are congruent.

Corollary 2 : Angle inscribed in a semicircle is a right angle.

2. Theorem of cyclic quadrilateral: Opposite angles of cyclic quadrilateral are supplementary.

Corollary of cyclic quadrilateral theorem: An exterior angle of cyclic quadrilateral is congruent to the angle opposite to its adjacent interior angle. S

3. Converse of cyclic quadrilateral theorem: If a pair of opposite angles of a quadrilateral is supplementary, then the quadrilateral is cyclic.

(4) If two points on a given line subtend equal angles at two distinct points which lie on the same side of the line, then the four points are concyclic.

(5) In the figure, chords AB and CD intersect at point E inside the circle, then ∠AEC = ∠DEB

= \(\frac{1}{2}\)[m(arc AC) + m(arc DB)] and

∠AED = ∠CEB = \(\frac{1}{2}\) [m(arc AD) + m(arc CB)].

(6) In the figure, line AB and CD intersect at point E outside the circle, then

∠BED = \(\frac{1}{2}\) [m(arc BD) – m(arc AC)].

**Theorem of angle between and tangent and secant:**

Statement : If an angle has its vertex on the circle, its one side touches the circle and the other intersects the circle in one more point, then the measure of the angle is half the measure of its Intercepted arc.

Given: Let O be the centre of the circle. Line DBC is tangent to the circle at point B. Seg BA is a chord of the circle. Point X of the circle is on C side of line BA and point Y of the circle is on D side of line BA.

To Prove: m∠ABC = \(\frac{1}{2}\) m(arc AXB).

Proof: Considering ∠ABC and the arc AXB intercepted by it, we see from the figures that arc AXB may be a semicircle, may be a minor arc or may be a major arc. We deal with these cases separately.

(a) When ∠ABC intercepts a semicircle, [as in figure (a)] the chord BA passes through the centre.

In that case, ∠ABC = 90° … (Tangent theorem)

Also since m(arc BXA) = 180°

∴ \(\frac{1}{2}\)m(arcBXA)= \(\frac{1}{2}\) × 180° = 90°

∴ m∠ABC = \(\frac{1}{2}\)m(arcAXB)

(b) When ∠ ABC intercepts a minor arc, [as in figure (b)] the centre lies in the exterior of ∠ABC.

We have, m∠ABC = 900_ m∠ABO

∴ m∠ABO = 90° – m∠ABC.

∴ m∠BAO = 90° – m∠ABC …. ( ABC is isosceles)

m∠ABO + m∠BAO = 180° – 2m∠ABC,

∴ 180° – m ∠BOA = 180° – 2m∠ABC

∴ m ∠BOA = 2m ∠ABC.

But m∠BOA = \(\frac{1}{2}\)m(arc AXB)

(c) When ∠ABC intercepts a major arc [as in figure (c)] the centre lies in the interior of ∠ABC.

We then have ∠ DBA intercepting a minor arc BYA.

**Alternative statement of the above theorem :**

In the figure, line PQ is a secant and line QR is tangent. The arc PSQ is intercepted by ∠ PQR. Chord PQ divides the circle in two parts. These are opposite arcs of each other. Now take any point T on the arc opposite to arc PSQ.

Here ∠ PQR = \(\frac{1}{2}\) m (arc PSQ) = ∠ PTQ.

**Converse of theorem of the angle between tangent and secant:**

Statement: A line is drawn from one end point of a chord of a circle and If the angle between the chord and the line is half the measure of the arc intercepted by that C angle, then that line Is a tangent to the circle.

In the figure,

if ∠ABC = \(\frac{1}{2}\) m(arc AXB) or ∠ABD = \(\frac{1}{2}\) m∠AYB then line DC is tangent to the circle.

[Note: Phis property is used in constructing a tangent to the given circle.]

**Theorem of Internal division of chords**

Statement: Suppose two chords of a circle intersect each other in the interior of the circle, then the product of the lengths of the two segments of one chord is equal to the product of the segments of the other chord.

GIven:

- A circle with centre O.
- Chords PR and QS intersect at point E inside the circle.

To prove: PE × ER = QE × ES.

Construction: Draw seg PQ and seg RS.

Proof:

**Theorem of external division of chords:**

Statement: If secants containing chords AB and CD of a circle intersect outside the circle in point E, then AE × EB = CE × ED.

Given:

- A circle with centre O.
- Secants AB and CD intersect at point E outside the circle.

To prove: AE × EB = CE × DE;

Construction: Draw seg AD and seg BC.

Proof:

**Tangent secant segments theorem:**

Statement: Point E is in the exterior of a cic1e. A secant through E intersects the circle at pointsA and B, and a tangent through E touches the circle at point T, then EA × EB = ET^{2}.

Given :

- A circle with centre O.
- Tangent ET touches the circle at point T.
- Secant EAB intersects the circle at points A and B.

To prove : EA × EB = ET^{2}.

Construction : Draw seg TA and seg TB.

Proof: In ∆ETA and ∆EBT,

**Remember this!**

**Theorem of internal division of chords.**

1.

In the figure, chord AB and CD intersect at point E Inside the circle then AE ×EB = CE × ED.

**Theorem of external division of chords.**

2.

In the figure, chord AB and chord CD intersect at point E outside the circle then

AE × EB = CE × ED.

**Tangent secant segments theorem.**

3.

In the figure, ray ET is a tangent at point T. Line EAB is secant intersecting the circle at points A and B then ET^{2} = EA × EB.