Definitions [5]
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.
Theorems and Laws [5]
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:
Proof:
-
Since XY is a tangent, it touches the circle only at P.
-
Point Q lies on the tangent XY and Q ≠ P, so Q lies outside the circle.
- Therefore, the distance OQ is greater than the radius OP.
OQ > OP -
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. - The shortest distance from a point to a line is perpendicular to the line.
Therefore, OP⊥XY
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:
-
Join OP, OQ and OR.
-
Radius is perpendicular to the tangent at the point of contact, so
∠OQP = ∠ORP = 90∘ -
OQ = OR (radii of the same circle).
-
OP = OP (common).
-
Therefore, △OQP ≅ △ORP (RHS).
- Hence, PQ = PR
Prove the following theorem:
Angles inscribed in the same arc are congruent.

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)]
Chords Intersecting Inside:
AE × EB = CE × ED
Chords Intersecting Outside:
AE × EB = CE × ED
Tangent–secant segments:
EA × EB = ET2
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.
Key Points
-
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
-
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°
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
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.
Important Questions [13]
- Choose the Correct Alternative: If the Points, A, B, C Are Non-collinear Points, Then How Many Circles Can Be Drawn Which Passes Through Points A, B, and C?
- In the Above Figure, the Circles with P, Q, and R Intersect at Points B, C, D, and E as Shown. Lines Cb and Ed Intersect in Point M. Lines Are Drawn from Point M to Touch
- In the given figure, O is centre of circle. ∠QPR = 70° and m(arc PYR) = 160°, then find the value of the following ∠QOR.
- In the given figure, O is centre of circle, ∠QPR = 70° and m(arc PYR) = 160°, then find the value of the following ∠PQR.
- In the given figure, O is the centre of the circle, ∠QPR = 70° and m(arc PYR) = 160°, then find the value of the following m(arc QXR).
- In the above figure, ∠L = 35°, find : m(arc MN) m(arc MLN) Solution : ∠L = 12 m(arc MN) ............(By inscribed angle theorem)∴ □=12 m(arc MN)∴ 2 × 35 = m(arc MN)
- In the above figure, chord PQ and chord RS intersect each other at point T. If ∠STQ = 58° and ∠PSR = 24°, then complete the following activity to verify: ∠STQ = 12 [m(arc PR) + m(arc SQ)] Activity
- Prove the following theorem: Angles inscribed in the same arc are congruent.
- In the following figure, m(arc PMQ) = 130o, find ∠PQS.
- In the following figure, secants containing chords RS and PQ of a circle intersects each other in point A in the exterior of a circle if m(arc PCR) = 26°, m(arc QDS) = 48°, then find
- If the Length of an Arc of the Sector of a Circle is 20 Cm and If the Radius is 7 Cm, Find the Area of the Sector.
- In the Following Figure, O is the Centre of the Circle. ∠Abc is Inscribed in Arc Abc And ∠ Abc = 65°. Complete the Following Activity to Find the Measure
- In the above figure, ∠ABC is inscribed in arc ABC. If ∠ABC = 60°. find m ∠AOC. Solution: ∠ABC = 12 m(arc AXC) ......□ 60° = 12 m(arc AXC) □ = m(arc AXC) But m ∠AOC = m(arc....) ......(Property of
