CBSE Class 10 Mathematics Circles Notes Set 04

Circle, Secant and Tangent

A circle is the locus of a point which moves in a plane in such a way that its distance from a fixed point remains constant. Consider a circle and a line AB. There are three different situations, as shown in the figure.

In figure (i) the line AB and the circle have no common point. In figure (ii) the line AB meets the circle at the two points P and Q; such a line is called a secant of the circle. In figure (iii) the line touches the circle at only one point P; such a line is called a tangent of the circle, and P is called the point of contact.

Secant: A line which intersects a circle at two distinct points is called a secant of the circle.
Tangent: A line which meets a circle at only one point is called a tangent to the circle. The point where the line meets the circle is called the point of contact. There is only one tangent passing through a point lying on the circle.

Theorem 1

Statement: A tangent to a circle is perpendicular to the radius through the point of contact.

Given: A circle with centre O and a tangent AB at the point P.
To prove: OP ⊥ AB.
Construction: Take any point Q, other than P, on the tangent AB. Join OQ. Suppose OQ meets the circle at R.
Proof: Among all line segments joining the point O to a point on AB, the shortest one is perpendicular to AB. So, to prove that OP ⊥ AB, it is sufficient to prove that OP is shorter than any other segment joining O to any point of AB.
Clearly OP = OR (radius). Now OQ = OR + RQ, so OQ > OR, that is OQ > OP (since OP = OR).
Thus OP is shorter than any other segment joining O to a point of AB. Hence OP ⊥ AB.

Theorem 2 (Converse of Theorem 1)

Statement: A line drawn through the end of a radius and perpendicular to it is a tangent to the circle.

Given: OP is a radius of a circle with centre O. AB is a line through P and OP ⊥ AB.
To prove: AB is a tangent to the circle at the point P.
Construction: Take a point R, other than P, on AB. Join OR. Suppose OR intersects the circle at Q.
Proof: OP ⊥ AB (given), so OP is the shortest line segment drawn from O to AB. Hence OP < OR, which gives OR > OP. Therefore R lies outside the circle. Thus every point on AB, other than P, lies outside the circle. This shows that AB meets the circle only at the point P. Hence AB is the tangent to the circle at P.

Number of Tangents from a Point to a Circle

(a) If a point is inside the circle, it is not possible to draw any tangent to the circle through this point, as shown in figure (i).
(b) If a point is on the circle, only one tangent to the circle can be drawn through this point, as shown in figure (ii).
(c) If a point is outside the circle, exactly two tangents can be drawn to the circle through this point, as shown in figure (iii). PA and PB are the lengths of the tangents drawn from P to the circle.

Theorem 3

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

Given: AP and AQ are two tangents drawn from a point A to a circle with centre O.
To prove: AP = AQ.
Construction: Join OP, OQ and OA.
Proof: In △AOQ and △AOP:
∠OQA = ∠OPA = 90° [a tangent is perpendicular to the radius through the point of contact]
AO = AO [common]
OQ = OP [radii]
So, by the R.H.S. criterion of congruence, △AOQ ≅ △AOP. Therefore AQ = AP [by CPCT]. Hence proved.

Results:
1. If two tangents are drawn to a circle from an external point, they subtend equal angles at the centre: ∠AOQ = ∠AOP [by CPCT].
2. If two tangents are drawn to a circle from an external point, they are equally inclined to the segment joining the centre to that point: ∠OAQ = ∠OAP [by CPCT].

Solved Examples

Check Your Progress

Check Your Progress 1. A quadrilateral PQRS is drawn to circumscribe a circle. If PQ = 12 cm, QR = 15 cm and RS = 14 cm, find the length of PS.
Answer:
In a quadrilateral circumscribing a circle, the sums of opposite sides are equal: PQ + RS = QR + PS.
12 + 14 = 15 + PS, so PS = 26 − 15 = 11 cm.
In simple words: The two sums of opposite sides must match.

Teacher's Note:
1. Use the property of a circumscribed quadrilateral (AB + CD = BC + DA).
2. Subtract the known side from the total.

Check Your Progress 2. O is the centre of a circle and PA and PB are tangents. If ∠PAB = 70°, find ∠APB.
Answer:
PA = PB (tangents from an external point), so △PAB is isosceles and ∠PBA = ∠PAB = 70°.
∠APB = 180° − 70° − 70° = 40°.
In simple words: The tangents are equal, so the base angles are equal.

Teacher's Note:
1. Tangents from one point are equal in length.
2. Then apply the angle sum of a triangle.

Check Your Progress 3. A tangent is drawn to a circle from a point 17 cm away from its centre. If the length of the tangent is 15 cm, find the radius of the circle.
Answer:
The radius through the point of contact is perpendicular to the tangent, so \( r^2 + 15^2 = 17^2 \).
\( r^2 = 289 - 225 = 64 \), so the radius is 8 cm.
In simple words: The radius, tangent and line to the centre form a right-angled triangle.

Teacher's Note:
1. Right angle is at the point of contact.
2. Hypotenuse is the distance from the centre.

Check Your Progress 4. Two concentric circles are of radii 5 cm and 3 cm. Find the length of the chord of the larger circle which is a tangent to the smaller circle.
Answer:
Half the chord = \( \sqrt{5^2 - 3^2} = 4 \) cm (the radius of the smaller circle is perpendicular to the chord at its midpoint).
Chord = 2 × 4 = 8 cm.
In simple words: Half the chord is a leg of a 3-4-5 right triangle.

Teacher's Note:
1. The perpendicular from the centre bisects the chord.
2. Double the half-chord.