A circle is a closed curve formed by all the points within a plane, which are all equidistant from any fixed central point. Therefore, some of the very important elements of a circle include the centre, radius, diameter, chord, and tangents.
This chapter centres on the properties and applications of tangents to a circle. These concepts are crucial to be studied not only for the purpose of board exams but also for higher studies in geometry. In order to help students grasp the material, these notes provide a summary of the chapter, including key ideas, definitions, formulae, and tips and tricks, in addition to solved examples.
Note: Tangent is a special case of Secant when the two endpoints of its corresponding chord coincide.
Theorem 1: The tangent at any point of a circle is perpendicular to the radius through the point of contact.
To prove: OP⟂ XY.
Given: XY is a tangent.
Construction: Mark a point Q on tangent XY. Join Q to O. That is the centre of the circle. Note that the Q must lie outside the circle.
Solution: we know by construction that Q is the point outside the circle (if the Q lies inside the circle, the XY will become a secant). Hence, OQ is longer than the radius OP as the radius lies inside. Mathematically, it can be written as,
OQ>OP
This will happen for every point on line XY except for the point P, which is the point of contact of the circle on tangent. Hence, OP is the shortest distance of all the distances between point O and the points of line XY. So, OP is Perpendicular to XY.
Theorem 2: The lengths of tangents drawn from an external point to a circle are equal.
To Prove: AP = BP
Given: AP and BP are tangents of a circle with centre O.
Construction: Joint O to P, A, and B.
Solution:
In APO and BPO
OP = OP (Common)
OA = OB (Radii of the same circle)
∠OAP = ∠OBP = 90 (the Radius of a circle is perpendicular to the tangent)
In APO ⩭ BPO (RHS)
AP = BP (CPCT)
Example 1: If d1, d2 (d2 > d1 ) are the diameters of two concentric circles and c is the length of a chord of a circle that is tangent to the other circle, prove that d22 = c2 + d12.
To Prove: d22 = c2 + d12
Given: AB = c.
Construction: Join O to B.
Solution: It is given that AB is the chord of the bigger circle.
Hence, OC bisects AB (the line through the centre of the circle bisects the chord of the circle)
BC = ½ AB = ½ c
Now, in triangle BCO by using Pythagorean theorem.
BO2 = OC2 + BC2
12d22=12d12+12c2
d22=d12+c2
or,
d12+c2=d22
Example 2: In Figure, from an external point P, a tangent PT and a line segment PAB are drawn to a circle with centre O. ON is perpendicular to the chord AB. Prove that : (i) PA.PB = PN2 – AN2 (ii) PN2 – AN2 = OP2 – OT2 (iii) PA.PB = PT2
Solution:
(i) PA can also be written as (PN – AN) similarly, PB = (PN + BN)
PA.PB = (PN – AN) (PN + BN)
AN = BN (line drawn from the centre of the circle to the chord bisects the cord)
= (PN – AN) (PN + AN)
(a2 – b2 = (a+b)(a–b)
= PN2 – AN2
(ii) As ON⊥PN Using Pythagoras theorem,
PN2 = OP2 – ON2
PN2 – AN2 = (OP2 – ON2) – AN2
= OP2 – (ON2 + AN2)
As ON⊥AN
OA2 = ON2 + AN2
= OP2 – OA2
OA = OT (radii of the same circle)
= OP2 – OT2
(iii) From (i) and (ii)
PA.PB = OP2 – OT2
∠OTP = 90° (radius of the circle makes a right angle with the tangent) hence,
PT2 = OP2– OT2
= PT2
Example 3: If a circle touches the side BC of a triangle ABC at P and extended sides AB and AC at Q and R, respectively, prove that AQ = ½(BC + CA + AB)
Solution:
(The lengths of tangents drawn from an external point to a circle are equal)
BQ = BP
CP = CR, and
AQ = AR
Now, 2AQ = AQ + AQ
2AQ = AQ + AR
AQ = AB + BQ, and AR = AC + CR
2AQ = (AB + BQ) + (AC + CR)
2AQ = AB + BP + AC + CP
2AQ = (BP + CP) + AC + AB
BC = BP + CP
2AQ = BC + CA + AB
AQ = ½(BC + CA + AB)
The CBSE Class 10 Maths Notes for Chapter 10 Circles are a priceless tool for understanding the material and performing well on tests because of their characteristics.
(Session 2025 - 26)