CBSE Class 10 Mathematics Circles Notes Set 03

Circle and Related Terms

A circle is a collection of all points in a plane, which are at a constant distance (radius) from a fixed point (centre) in that plane. If \( A_1, A_2, A_3, \ldots \) are points on the circle with centre C and radius r, then \( CA_1 = CA_2 = CA_3 = \ldots = r \). You have learnt some of the properties of a circle in Class IX, and you must be aware of various terms related to a circle such as chord, sector, segment, arc etc.

Chord is a line segment that joins two points on the circumference of the circle. A diameter is the longest chord of the circle and it passes through the centre. In the figure, BC is a diameter, DE is a chord and OA is a radius.
An arc is the portion of the circumference of a circle, for example the arc DE.
The region between a chord and either of its arcs is called a segment of the circle.
The part of a circle made of an arc along with its two radii is called a sector of the circle.

Secant and Tangent

A secant is a line that intersects a circle at two distinct points. A tangent to a circle is a line that touches the circle at exactly one point. The point at which the tangent line meets the circle is known as the point of contact. The length of the tangent is the length of the line segment of the tangent between a given point and the point of contact with the circle. In the figure, the line XY is a secant, the line QT is the tangent at the point of contact P, and the length of the tangent from a point Q on it is QP.

Note.
1. A tangent does not pass through a point lying inside the circle.
2. There is only one tangent that passes through a point lying on a circle.
3. Exactly two tangents can pass through a point lying outside a circle.

Properties of a Tangent

Theorem 1. 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 at a point P of the circle. To prove: \( OP \perp XY \).
Construction: Take a point Q, other than P, on XY and join OQ.
Proof: Q lies on the tangent XY, so Q lies outside the circle (the tangent meets the circle only at P). Let OQ meet the circle at M. Then \( OM < OQ \) ...(i), because a part is less than the whole. But \( OP = OM \) (radii of the same circle) ...(ii). From (i) and (ii), \( OP < OQ \). So OP is shorter than any other line segment drawn from O to a point of XY, i.e. OP is the shortest distance from O to the line XY. The shortest distance from a point to a line is the perpendicular distance. Therefore \( OP \perp XY \). Hence proved.

Theorem 2 (converse of Theorem 1). A line drawn through the end of a radius and perpendicular to the radius is a tangent to the circle.
Given: A circle with centre O having radius OP, and a line XY through P such that \( OP \perp XY \). To prove: XY is a tangent to the circle at P.
Proof: Take any point Q, other than P, on XY and join OQ. Since \( OP \perp XY \), OP is the shortest distance from O to XY, so \( OQ > OP \). As OP is the radius, Q lies outside the circle. Therefore every point of XY other than P lies outside the circle, i.e. XY meets the circle only at P. Hence XY is the tangent to the circle at the point P.

Theorem 3. The lengths of the tangents drawn from an external point to a circle are equal.
Given: Two tangents PQ and PR are drawn from a point P to a circle with centre O. To prove: \( PQ = PR \).
Construction: Join OP, OQ and OR.
Proof: PQ is a tangent at Q and OQ is the radius, so \( OQ \perp PQ \); similarly \( OR \perp PR \). In the right triangles OPQ and OPR: \( OQ = OR \) (radii of the same circle), \( OP = OP \) (common), and \( \angle OQP = \angle ORP = 90^\circ \). So \( \triangle OPQ \cong \triangle OPR \) (RHS congruence). Hence \( PQ = PR \) (C.P.C.T.).

Theorem 4. When two tangents are drawn from an external point to a circle, then (i) they form equal angles at the centre, i.e. \( \angle POQ = \angle POR \), and (ii) they are equally inclined to the line segment joining the centre to that point, i.e. \( \angle OPQ = \angle OPR \).
Proof: In \( \triangle POQ \) and \( \triangle POR \): \( PQ = PR \) (tangents from an external point), \( OQ = OR \) (radii), \( OP = OP \) (common). So \( \triangle POQ \cong \triangle POR \) (SSS). Hence \( \angle POQ = \angle POR \) and \( \angle OPQ = \angle OPR \) (C.P.C.T.).

Solved Examples

Example 1. A tangent QT of length 8 cm is drawn from a point Q, which is 10 cm away from the centre of a circle. Find the radius of the circle.

Answer:
Let O be the centre and T the point of contact, so \( OQ = 10 \) cm and \( QT = 8 \) cm. Join OT.
QT is a tangent at T and OT is the radius through T, so \( OT \perp QT \).
In the right \( \triangle OTQ \): \( OT = \sqrt{OQ^2 - QT^2} = \sqrt{10^2 - 8^2} = \sqrt{36} = 6 \) cm.
Hence the radius of the circle is 6 cm.

Example 2. In the given figure, \( PQ = QR \), \( \angle RQP = 68^\circ \), and PC and CQ are tangents to the circle with centre O. Calculate (i) \( \angle QOP \), (ii) \( \angle QCP \).

Answer:
(i) In \( \triangle PQR \), \( PQ = QR \Rightarrow \angle PRQ = \angle QPR \). Their sum is \( 180^\circ - 68^\circ = 112^\circ \), so \( \angle PRQ = 56^\circ \).
The angle subtended by an arc at the centre is double the angle subtended at any point of the remaining circle: \( \angle QOP = 2\angle PRQ = 2 \times 56^\circ = 112^\circ \).
(ii) The radius through the point of contact is perpendicular to the tangent, so \( \angle OQC = \angle OPC = 90^\circ \).
In the quadrilateral OPCQ: \( 90^\circ + 112^\circ + 90^\circ + \angle QCP = 360^\circ \Rightarrow \angle QCP = 68^\circ \).
So \( \angle QOP = 112^\circ \) and \( \angle QCP = 68^\circ \).

Example 3. Two circles touch each other externally at the point P. Q is a point on the common tangent through P. Prove that the tangents QA and QB are equal.

Answer:
QA and QP are tangents drawn from the external point Q to the first circle, so \( QA = QP \) ...(i)
QB and QP are tangents drawn from Q to the second circle, so \( QB = QP \) ...(ii)
From (i) and (ii), \( QA = QB \). Hence proved.

Example 4. From an external point M, two tangents MP and MQ are drawn to the circle with centre O. Prove that OM is the perpendicular bisector of PQ.

Answer:
Let OM intersect PQ at R. In \( \triangle MPR \) and \( \triangle MQR \): \( MP = MQ \) (tangents from an external point), \( \angle PMR = \angle QMR \) (the tangents are equally inclined to OM), and \( MR = MR \) (common).
So \( \triangle MPR \cong \triangle MQR \) (SAS). Hence \( PR = QR \) and \( \angle PRM = \angle QRM \) (C.P.C.T.).
But \( \angle PRM + \angle QRM = 180^\circ \) (linear pair), so \( \angle PRM = \angle QRM = 90^\circ \).
Thus \( PR = QR \) and \( OM \perp PQ \), i.e. OM is the perpendicular bisector of PQ.

Example 5. In two concentric circles, prove that all the chords of the outer circle which touch the inner circle are of equal length.

Answer:
Let O be the common centre. Let PQ and RS be two chords of the outer circle that touch the inner circle at M and N respectively.
The radius is perpendicular to the tangent at the point of contact: \( OM \perp PQ \) and \( ON \perp RS \).
The perpendicular from the centre to a chord bisects the chord, so \( PM = \frac{1}{2}PQ \) and \( RN = \frac{1}{2}RS \).
In the right \( \triangle OMP \) and \( \triangle ONR \): \( OM = ON \) (radii of the inner circle), \( OP = OR \) (radii of the outer circle), and \( \angle OMP = \angle ONR = 90^\circ \). So \( \triangle OMP \cong \triangle ONR \) (RHS), hence \( PM = RN \).
Therefore \( PQ = 2PM = 2RN = RS \). Hence all such chords are equal in length.

Example 6. In the given figure, a circle is inscribed in \( \triangle PQR \) and touches the sides PQ, PR and QR at M, N and S respectively. If \( SQ = SR \), prove that \( \angle PQR = \angle PRQ \).
Answer:
Tangents drawn from an external point are equal: \( QM = QS \) and \( RN = RS \).
Given \( QS = RS \), so \( QM = RN \) ...(i)
Also \( PM = PN \) (tangents from P) ...(ii)
Adding (i) and (ii): \( PM + QM = PN + RN \Rightarrow PQ = PR \).
So \( \triangle PQR \) is isosceles and \( \angle PQR = \angle PRQ \) (angles opposite equal sides). Hence proved.

Summary

1. A secant meets a circle in two points; a tangent meets it in exactly one point, called the point of contact.
2. A tangent cannot pass through a point inside the circle; exactly one tangent passes through a point on the circle; exactly two tangents pass through a point outside the circle.
3. The tangent at any point of a circle is perpendicular to the radius through the point of contact.
4. The lengths of the two tangents drawn from an external point to a circle are equal.