Official ICSE Book for Class 10 Mathematics: Chapter 21 Tangent Properties of Circles
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Chapter 21
Tangent Properties Of Circles
Points To Remember
1. Some Results (Theorems)
Theorem 1. The tangent at any point of a circle and the radius through the point are perpendicular to each other.
Given. A circle with centre O, AB is a tangent to the circle at a point P and OP is the radius through P.
To prove. OP - AB.
Construction. Take a point Q, other than P, on tangent AB. Join OQ.
Proof.
| Statement | Reason |
|---|---|
| 1. Since Q is a point on tangent AB, other than the point P, so Q will be outside the circle. - OQ will intersect the circle at some point R. | Tangent at P intersects the circle at point P only. |
| 2. - OR - OQ =- OP - OQ | A part is less than its whole. OR = OP = radius. |
| 3. Thus, OP is shorter than any other line segment joining O to any point of AB. | |
| 4. OP - AB | Of all line segments drawn from O to line AB, the perpendicular is the shortest. |
| Hence, the radius OP is perpendicular to tangent at P. | |
Theorem 2. If two circles touch each other, the point of contact lies on the straight line through their centres.
Case 1. When the given two circles touch each other externally.
Given. Two circles with centres A and B, touching each other externally at a point P.
To prove. P lies on line AB.
Construction. At the point P, draw a common tangent PT to the two circles. Join AP and BP.
Proof.
| Statement | Reason |
|---|---|
| 1. - APT = 90° | Radius through the point of contact is perpendicular to the tangent. |
| 2. - BPT = 90° | Same as above. |
| 3. - APT + - BPT = 180° =- - APB = 180° =- APB is a straight line. - P lies on line AB. | Adding 1 and 2. |
Case II. When the given two circles touch each other internally.
Given. Two circles with centres A and B, touching each other internally at a point P.
To prove. P lies on line AB.
Construction. At the point P, draw a common tangent PT. Join AP and BP.
Proof.
| Statement | Reason |
|---|---|
| 1. AP - PT | Radius through the point of contact is perpendicular to the tangent. |
| 2. BP - PT | Same as above. |
| 3. AP and BP are both perpendicular to the same line PT. | From 1 and 2. |
| 4. AP and BP lie in the same line =- ABP is a straight line. - P lies on line AB. |
Theorem 3. If two tangents are drawn to a circle from an exterior point, then (i) the tangents are equal in length; (ii) the tangents subtend equal angles at the centre; (iii) the tangents are equally inclined to the line joining the point and the centre of the circle.
Given. PA and PB are two tangents drawn to a circle with centre O, from an exterior point P.
To prove.
(i) PA = PB,
(ii) - AOP = - BOP,
(iii) - APO = - BPO.
Proof.
| Statement | Reason |
|---|---|
| 1. In - AOP and - BOP: OA = OB - OAP = - OBP = 90° OP = OP - - AOP = - BOP | Radii of the same circle. Radius through point of contact is perpendicular to the tangent. common. S.S.A. (axiom of congruency) |
| 2. Hence, we have (i) PA = PB (ii) - AOP = - BOP (iii) - APO = - BPO | c.p.c.t. c.p.c.t. c.p.c.t. |
Intersecting Chords And Tangents:
(a) Segments of a chord:
(i) If P is a point on a chord AB of a circle, then we say that P divides AB internally into two segments PA and PB.
(ii) If AB is a chord of a circle and P is a point on AB produced, we say that P divides AB externally into two segments PA and PB.
Alternate Segments:
In the given figure, PAT is a tangent to the circle at a point A and AB is a chord. The chord AB divides the circle into two segments, namely ADB and BCA, called the alternate segments.
For - BAT, the alternate segment is BCA.
For - BAP, the alternate segment is ADB.
Some More Results (Theorems)
Theorem 1. If two chords of a circle intersect internally or externally, then the products of the lengths of their segments are equal.
Case 1. When the two chords intersect internally.
Given. Two chords AB and CD of a circle intersect each other at a point P inside the circle.
To prove. PA x PB = PC x PD.
Construction. Join AC and BD.
Proof.
| Statement | Reason |
|---|---|
| 1. In - APC and - DPB, (i) - ZAC = - ZDP B (ii) - PAC = - PDB - - APC - - DPB | Vert. opp. angles angles in the same segment. By AA-similarity axiom. |
| 2. - PA/FD = PC/PB =- PA x PB = PC x PD. Thus, in this case PA x PB = PC x PD. | Corresponding sides of similar - s are proportional |
Case II. When the two chords intersect externally
Given. Two chords AB and CD of a circle, when produced, intersect each other at a point P, outside the circle.
To prove. PA x PB = PC x PD.
Construction. Join AC and BD.
Proof.
| Statement | Reason |
|---|---|
| 1. In - PDB and - PAC, (i) - ZDB = - ZAC (ii) - ZBD = - ZCA - - PDB - - PAC | Exterior angle of a cyclic quad. = Int. opp. angle Exterior angle of a cyclic quad. = Int. opp. angle By AA-similarity axiom. |
| 2. - PD/PA = PB/PC =- PA x PB = PC x PD. - In this case also, PA x PB = PC x PD. | Corresponding sides of similar - s are proportional |
Hence, in both the cases, we have PA x PB = PC x PD.
Teacher's Note
When two chords intersect inside a circle - like two crossing strings on a drum - the relationship PA x PB = PC x PD helps architects design circular structures with intersecting support beams.
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