Revision: Circle Geometry Maths 2 SSC (English Medium) 10th Standard Maharashtra State Board

A secant is a straight line that passes through the circle and intersects it at two distinct points. 

line AB cuts the circle at points M and N ⇒ AB is a secant.

A tangent is a straight line that touches a circle at exactly one point only, without cutting through it. This single point where the tangent touches the circle is called the point of contact or point of tangency.

Line CD touches the circle at point P only ⇒ CD is a tangent

Point P is the point of contact.

Two circles are touching if they intersect at exactly one point.

Types:

• Externally touching circles
  Distance between centres = sum of radii

• Internally touching circles
  Distance between centres = difference of radii

An inscribed angle is an angle whose vertex lies on the circle and whose arms intersect the circle at two other distinct points.

The arc of the circle intercepted by the arms of the angle is called the intercepted arc of the inscribed angle.

Statement: The lengths of tangents drawn from an external point to a circle are equal.

Given: A circle with centre O and two tangents PQ and PR drawn from an external point P.

To Prove: PQ = PR

Proof:

1. Join OP, OQ and OR.

2. Radius is perpendicular to the tangent at the point of contact, so

   ∠OQP = ∠ORP = 90^∘

3. OQ = OR (radii of the same circle).

4. OP = OP (common).

5. Therefore, △OQP ≅ △ORP (RHS).

6. Hence, PQ = PR

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

Given: A circle with centre O and a tangent XY touching the circle at P.

To Prove: $OP\perp XY$

Proof:

1. Since XY is a tangent, it touches the circle only at P.

2. Point Q lies on the tangent XY and Q ≠ P, so Q lies outside the circle.

3. Therefore, the distance OQ is greater than the radius OP.
   OQ > OP

4. This is true for every point Q on the line XY except P.
   Hence, OP is the shortest distance from O to the line XY.

5. The shortest distance from a point to a line is perpendicular to the line.
   Therefore, OP⊥XY

m∠STQ = m∠SPQ + \[\boxed{\text{m∠PSR}}\]   ...[Theorem of the external angle of a triangle]

 = \[\frac{1}{2} \text{m(arc SQ)}\] + \[\boxed{\frac{1}{2} \text{m(arc PR)}}\]   ...[Inscribed angle theorem]

= \[\frac{1}{2} [\boxed{\text{m(arc PR)}} + \boxed{\text{m(arc SQ)}}]\]

Proof: Draw seg GF.

∠EFG = ∠FGH   ...\[\boxed{\text{[Alternate angles]}}\] (I)

∠EFG = \[\boxed{\frac{1}{2} (\text{arc EG)}}\]   ...[Inscribed angle theorem] (II)

∠FGH = \[\boxed{\frac{1}{2} (\text{arc FH)}}\]   ...[Inscribed angle theorem] (III)

∴ m(arc EG) = \[\boxed{\text{m(arc FH)}}\]   ...[By (I), (II) and (III)]

chord EG ≅ chord FH   ...[Corresponding chords of congruent arcs]

Proof: Segment AC is a diameter of the circle.

∴ m(arc AXC) = \[\boxed{180°}\]   ...(i) [Measure of semi circular arc is 180°]

Arc AXC is intercepted by the inscribed angle ∠ABC

∠ABC = \[\boxed{\frac{1}{2} \text{m(arc AXC)}}\]   ...[Inscribed angle theorem]

= \[\frac{1}{2} \times \boxed{180°}\]   ...[From (i)]

∴ m∠ABC = \[\boxed{90°}\]

∴ ∠ABC is a right angle.

Proof:

`m∠PQR = 1/2 xx [m(arc  PTR)]`   ...(i) \[\boxed{\text{Inscribed angle theorem}}\]

m∠\[\boxed{PSR}\] = `1/2 xx [m(arc  PTR)]`   ...(ii) \[\boxed{\text{Inscribed angle theorem}}\]

m∠\[\boxed{\text{PQR}}\] = m∠PSR   ...[By (i) and (ii)]

∴ ∠PQR ≅ ∠PSR

∆ABC is an equilateral triangle.

∴ ∠ABC = ∠ACB = ∠BAC = 60°   ...(i) [Angles of an equilateral triangle]

`∠CBP = 1/2 ∠ABC`   ...[Ray BP bisects ∠B]

∴ `∠CBP = 1/2 xx 60^circ`   ...[From (i)]

∴ ∠CBP = 30°

∠CBP = ∠CAP = 30°   ...[Angles inscribed in the same arc]

∴ ∠CAQ = 30°   ...(ii) [A–P–Q]

In ∆ABQ,

∠BAQ = ∠BAC + ∠CAQ   ...[Angle addition property]

∴ ∠BAQ = 60° + 30°   ...[From (i) and (ii)]

∴ ∠BAQ = 90°

Also, ∠ABQ = 60°   ...[From (i) and B–C–Q]

∴ ∠BQA = 30°   ...[Remaining angle of ∆ABQ]

∴ ∠CQA = 30°   ...(iii) [B–C–Q]

In ∆CQA,

∠CAQ = ∠CQA   ...[From (ii) and (iii)]

∴ CQ = CA   ...[Converse of isosceles triangle theorem]

Proof: 

∠ACB = \[\boxed{90°}\]   ...[Angle inscribed in semicircle]

∠DCB = \[\boxed{\text{∠DCA} = 45°}\]   ...[CD is the bisector of ∠C]

m(arc DB) = \[\boxed{\text{2∠DCB} = 90°}\]   ...[Inscribed angle theorem]

∠DOB = \[\boxed{\text{m(arc DB)} = 90°}\]   ...[Definition of measure of arc] (i)

seg OA ≅ seg OB   ...\[\boxed{\text{[Radii of the same circle]}}\] (ii)

∴ Line OD is \[\boxed{\text{perpendicular bisector}}\] of seg AB   ...[From (i) and (ii)]

∴ seg AD ≅ seg BD

Construction: Draw seg EO, seg FO, seg GO, seg HO, seg GF.

Proof:

Chord EF || chord GH, GF is the transversal

∠EFG ≅ ∠HGF   ...(i) [Alternate angles]

`{:(∠EFG = 1/2 m(arc  GE)),(∠HGF = 1/2 m(arc  HF)):}}`   `{:(...("ii")),(...["Inscribed angle theorem"]),(...("iii")):}`

∴ m(arc GE) = m(arc HF)   ...(iv) [From (i), (ii) and (iii)]

`{:("Also"","  ∠EOG = m(arc  GE)),(∠FOH = m(arc  HF)):}}`   `{:(...("v") ["Definition of measure of arc"]),(...("vi")):}`

∴ ∠EOG = ∠FOH   ...[From (iv), (v) and (vi)]

Given: C is the circle's center. In the same arc PTR, ∠PQR and ∠PSR are written.

To prove: ∠PQR ≅ ∠PSR

Proof:

arc PTR is intercepted by ∠PQR.

arc PTR is intercepted by ∠PSR.

`∠PQR = 1/2 m(arc  PTR)`   ...(i) [Inscribed angle theorem]

And 

`∠PSR = 1/2 m(arc  PTR)`   ...(ii) [Inscribed angle theorem]

∴ ∠PQR ≅ ∠PSR   ...[From (i) and (ii)] 

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

Measure of an inscribed angle = `1/2` × measure of intercepted arc

• Angles inscribed in the same arc are congruent.
• An angle inscribed in a semicircle is a right angle.

Chords Intersecting Inside:

AE × EB = CE × ED

Chords Intersecting Outside:

AE × EB = CE × ED

Tangent–secant segments:

EA × EB = ET²

• A tangent touches a circle at only one point (point of contact).

• The radius through the point of contact is perpendicular to the tangent.

• A line perpendicular to the radius at its endpoint is a tangent to the circle.

• No tangent can be drawn to a circle from a point inside the circle.

• Exactly one tangent can be drawn from a point on the circle.

• Exactly two tangents can be drawn from a point outside the circle.

• From an external point, the two tangents drawn to a circle are equal in length.

• The two tangents from an external point make equal angles at the centre.

• If two circles touch each other, the point of contact lies on the line joining their centres (external and internal touching).

A cyclic quadrilateral: All four vertices lie on the same circle

Key properties:

• Opposite angles are supplementary

• Exterior angle = interior opposite angle

Two arcs are congruent if:

• They belong to the same or congruent circles

• They have equal radii

• They have equal measures

Congruent arcs ⇔ congruent chords

• An angle whose vertex is the centre of a circle is called a central angle.

• Measure of minor arc = measure of its central angle

• Measure of major arc = 360° − minor arc

• The measure of a semicircle is 180°

• Full circle = 360°

1) Chords intersect inside the circle

Angle = `1/2` (sum of intercepted arcs)

(2) Secants intersect outside the circle

Angle = `1/2`(difference of intercepted arcs)

(3) Tangent–secant angle

Angle = `1/2` (intercepted arc)

These concepts unify angles + arcs.