Important Results On Tangents To A Circle

Important Results on Tangents to a Circle Tangent: A line that intersects the circle in exactly one point is called a tangent to the circle. The point at which the tangent intersects the circle is called the point of content. In figure 8-41, ST is the tangent and P is the point of contact. It is also called the point of tangency.

The following points should be noted carefully 1. From a point, in the interior of a circle, no tangent can be drawn to the circle. 2. From a point on the circle, a unique tangent can be drawn to the circle. The tangent passes through that point. 3. From a point in the exterior of a circle, two tangents can be drawn to the circle. 4. Of all the points on a tangent to a circle, the point of contact is the nearest to the centre of the circle. Theorem: A tangent at any point of a circle is perpendicular to the vertex through the point of contact. Given: A circle with centre O. PT is a tangent line to the circle at point P.

To prove: OP ⊥ PT . Construction: Let Q be any point other than P on PT. Join OQ. Proof:

Since Q is any point other than P on PT.

Therefore Q lies in the exterior of circle OQ>OP. This shows that of all segments that can be drawn from the centre O to any point on the line PT, OP is the shortest. We know that the shortest segment that can be drawn from a given point to a given line is the perpendicular from the given point to the given line Hence, OP ⊥ PT . Theorem: The lengths of two tangents drawn from an external point to a circle are equal. Given: AP and AQ are two tangents from a point A to a circle C ( O, r ) To prove: AP = AQ Construction: Join OA, OP and OQ

Proof: Since a tangent is perpendicular to the radius through the point of contact

Therefore OP ⊥ AP

and OQ ⊥ AQ .

Now in right triangles Triangle OAP and Triangle OAQ , we have OP = OQ

(Radii of the same circle)

∠OPA = ∠OQA

( Each = 90o )

OA = OA

(Common to both the triangles)

Triangle OAP ≅ Triangle OAQ Thus AP = AQ

(RHS criterion of congruency)

(cpctc)

Theorem: If two chords of a circle intersect inside or outside the circle when produced, the rectangle formed by the two segments of one chord is equal in area to the rectangle formed by the two segments of another chord.

Given: Two chords AB and CD of a circle such that they interest each other at a point P, lying inside or outside the circle. To prove: PA. PB = PC. PD Construction: Join AC and BD Proof: Case I: Point P lies inside the circle In triangles Triangle PCA and Triangle PBD , we have ∠PCA = ∠PBD

(Angles in the same segment are equal)

∠APC = ∠BPD

(Vertically opposite angles are equal)

Therefore Triangle PCA similar to Triangle PBD (AA similarity criterion) So,

PA PC = PD PB

(Property of similar triangles)

Thus PA. PB = PC. PD Case II:

Point P lies outside the circle

Since angles ∠PAC and ∠CAB form a linear pair

∠PAC + ∠CAB = 180o and ∠CAB + ∠PDB = 180o (Opposite angles of a cyclic

quadrilateral are supplementary)

Therefore ∠PAC = ∠PDB

… (1)

In triangles PCA and PBD, ∠PAC = ∠PDB

[from (1)]

∠APC = ∠DPB

(same angle)

Therefore Triangle PCA similar to Triangle PBD So,

PA PC = PD PB

(AA similarity criterion)

(Property of similar triangles)

Thus PA. PB = PC. PD Hence, in either case PA. PB = PC. PD Theorem: (Converse of Previous Theorem): If two straight lines intersect internally or externally such that the rectangle contained by the parts of one is equal to the rectangle contained by the parts of the other, the four extremities of the two straight lines are con-cyclic.

Given: Two straight lines AB and CD intersecting at P so that PA × PB = PC × PD To prove: A, B, C and D are con-cyclic

Construction: Draw a circle passing through A, B and C. If it does not pass through D, suppose it cuts CD in D ' . Proof: Since AB and CD ' are two chords of circle intersecting each other at P, therefore PA × PB = PC × PD ' (by theorem) … (1) But PA × PB = PC × PD

(given)

Therefore PC × PD ' = PC × PD

[from (1) and (2)]

… (2)

Thus, PD ' = PD But this is possible only if D ' coincides with D. So, the circle passes through D. Hence A, B, C and D are con-cyclic Theorem: If PAB is secant to a circle intersecting it at A and B, and PT is a tangent, then PA.PB = PT 2

Given: A secant PAB to a circle with centre O intersecting it at A and B and a tangent PT to the circle. To prove: PA.PB = PT 2

Construction: Join O to the mid point M of AB and also to A, P and T. Proof: PA = PM – AM PB = PM + MB PB = PM + AM

( Since AM = MB )

Therefore, PA.PB = ( PM − AM )( PM + AM ) or PA.PB = PM 2 − AM 2

[ (a − b)(a + b) = a 2 − b 2 ]… (1)

Again OM is perpendicular to AB Therefore PM 2 = OP 2 − OM 2 and AM 2 = OA2 − OM 2

(Pythagoras theorem) (Pythagoras theorem)

Substituting the values of PM 2 and AM 2 in (1), we have PA.PB = (OP 2 − OM 2 ) − (OA2 − OM 2 ) PA.PB = OP 2 − OM 2 − OA2 + OM 2 PA.PB = OP 2 − OA2

PA.PB = OP 2 − OT 2 ( Since OA = OT , being radii of the same circle) PA.PB = PT 2

(By Pythagoras theorem)

Theorem: If a line touches a circle and from the point of contact a chord is drawn, the angles which this chord makes with the given line are equal respectively to the angles formed in the corresponding alternate segments. Given: PQ is a tangent to a circle with centre O at a point A, AB is a chord and C, D are points in the two segments of the circle formed by the chord AB. To prove: ∠BAQ = ∠ACB and ∠BAP = ∠ADB Construction: Draw the diameter AOE and join EB.

Proof: In AEB

∠ABE = 90o

(Angle in a semi-circle is right angle)

Therefore ∠AEB + ∠EAB = 90o

… (1)

Now, EA ⊥ PQ,

Therefore ∠EAB + ∠BAQ = ∠EAQ = 90o

… (2)

From (1) and (2), we have

∠AEB = ∠BAQ

(Each = 90o − ∠EAB )

Also, ∠ACB = ∠AEB

(Angles in the same segment)

Therefore ∠BAQ = ∠ACB

… (3)

Also, ∠BAQ + ∠BAP = 180o

(linear pair)

and ∠ACB + ∠ADB = 180o

(Opposite angles of a cyclic quadrilateral)

Therefore, ∠BAQ + ∠BAP = ∠ACB + ∠ADB Using (3), we get ∠BAP = ∠ADB

Theorem: (Converse of Previous Theorem). If a line is drawn through an end point of a chord of a circle so that the angle formed by it with the chord is equal to the angle subtended by the chord in the alternate segment, then the line is tangent to the circle.

Given: AB is a chord of a circle and a line PAQ such that ∠BAQ = ∠ACB , where C is any point in the alternate segment ACB. To prove: Line PAQ is a tangent to the circle Construction: If PAQ is not the tangent to the circle, draw tangent P ' AQ ' at A. Proof: Since P ' AQ ' is a tangent and AB is a chord of the circle, therefore, by alternate segment theorem.

∠BAQ ' = ∠ACB But ∠BAQ = ∠ACB

(Given)

Therefore ∠BAQ ' = ∠BAQ Hence, AQ ' coincides with AQ or in other words P ' AQ ' coincides with PAQ. This shows that PAQ is the tangent to the circle at A.

Theorem: If two circles touch each other internally or externally, the point of contact lies on the line joining their centers. Given: Two circles with centres A and B touching each other (i) internally and (ii) externally at P. To prove: P lies on AB Construction: Draw PT, the common tangent, at P to both the circles.

Proof: Since BP is a radius and PT is a tangent

Therefore ∠BPT = 90o contact)

(Radius is perpendicular to the tangent at the point of

Similarly, ∠APT = 90o In figure 8-49 (i), since ∠APT = ∠BPT = 90o Therefore their arms AP and BP coincide. Hence B, A, P lie in a straight line. In figure 8-49 (ii), since ∠BPT + ∠APT = 180o

( Since ∠BPT = ∠APT = 90o )

Therefore APB is a straight line Hence, points A, B and P are in the same straight line Remarks: 1. If two circles touch internally, the distance between their centers is equal to the difference of their radii. 2. If they touch externally, the distance between their centers is equal to the sum of their radii.

**************************************************