NCERT Exemplar Solutions - Class 10 - Mathematics - Unit 9: Circles

Exercise 9.2

Step 1: We have two circles with the same center (concentric). Outer circle radius = \(R = 5\,\text{cm}\). A chord \(AC\) of the outer circle has length 8 cm, and this chord just touches (is tangent to) the inner circle.

Step 2: If a chord of the outer circle is at a distance \(d\) from the center, then the formula for the chord length is: \[ L = 2\sqrt{R^2 - d^2} \] where: • \(L\) = chord length, • \(R\) = radius of outer circle, • \(d\) = perpendicular distance from center to chord.

Step 3: In our case, the distance \(d\) is also the radius of the inner circle (because the chord is tangent to the inner circle). So, let inner radius = \(r\). That means \(d = r\).

Step 4: Substitute the known values into the chord formula: \[ 8 = 2\sqrt{5^2 - r^2} \]

Step 5: Simplify step by step: \[ 8 = 2\sqrt{25 - r^2} \] Divide both sides by 2: \[ 4 = \sqrt{25 - r^2} \]

Step 6: Square both sides to remove the square root: \[ 4^2 = 25 - r^2 \] \[ 16 = 25 - r^2 \]

Step 7: Rearrange to find \(r^2\): \[ r^2 = 25 - 16 = 9 \]

Step 8: Take square root: \[ r = \sqrt{9} = 3 \]

Final Answer: The radius of the inner circle is \(3\,\text{cm}\).

Step 1: We know that the radius of a circle is always perpendicular to the tangent at the point of contact.

So, \(OQ \perp PQ\) and \(OR \perp PR\).

Step 2: This means angle \(OQP = 90^\circ\) and angle \(ORP = 90^\circ\).

Step 3: Now, look at quadrilateral \(QORP\). Its vertices are \(Q, O, R, P\).

In this quadrilateral, the opposite angles are:

• At vertex \(Q\): \(\angle OQP = 90^\circ\)
• At vertex \(R\): \(\angle ORP = 90^\circ\)

Step 4: Add these opposite angles:

\(90^\circ + 90^\circ = 180^\circ\).

Step 5: A property of a cyclic quadrilateral is: If a pair of opposite angles is supplementary (i.e., adds up to \(180^\circ\)), then the quadrilateral is cyclic.

Step 6: Since in quadrilateral \(QORP\), one pair of opposite angles is supplementary, we can say that \(QORP\) is a cyclic quadrilateral.

Step 1: Join \(OC\) and \(OD\), where \(O\) is the centre of the circle and \(C, D\) are the points of tangency.

Step 2: A radius drawn to the point of contact is always perpendicular to the tangent. So, \(OC \perp BC\) and \(OD \perp BD\). That means \(\angle OCB = 90^\circ\) and \(\angle ODB = 90^\circ\).

Step 3: The angle between the tangents at \(B\) is given as \(\angle DBC = 120^\circ\). The angle at the centre, \(\angle COD\), is supplementary to this (they add up to \(180^\circ\)). So, \(\angle COD = 180^\circ - 120^\circ = 60^\circ\).

Step 4: Tangents drawn from an external point are equal in length. Hence, \(BC = BD\). Because of this symmetry, the line \(OB\) will bisect the angle between \(BC\) and \(BD\). So, \(\angle OBC = 60^\circ\).

Step 5: Now consider right triangle \(\triangle OBC\). Its angles are: - \(\angle OCB = 90^\circ\) (because radius is perpendicular to tangent), - \(\angle OBC = 60^\circ\), - So, \(\angle BOC = 30^\circ\).

Step 6: In a right triangle with angles \(30^\circ, 60^\circ, 90^\circ\), the sides are in the ratio: \(1 : \sqrt{3} : 2\). - Side opposite \(30^\circ = 1\), - Side opposite \(60^\circ = \sqrt{3}\), - Hypotenuse = 2.

Step 7: In \(\triangle OBC\): - \(BC\) is opposite \(30^\circ\), - \(OB\) is the hypotenuse. Therefore, \(OB = 2 \times BC\).

Step 8: Since \(BC = BD\), we get: \(BC + BD = BC + BC = 2BC = OB\).

Hence proved: \(BC + BD = BO\).

Step 1: Let the two straight lines intersect at a point \(X\).

Step 2: Suppose a circle is drawn that touches both lines. Let the centre of the circle be \(O\), and let it touch the first line at point \(T_1\) and the second line at point \(T_2\).

Step 3: A basic property of circles is: the radius drawn to the point of contact with a tangent is perpendicular to that tangent.

So, \(OT_1 \perp \text{first line}\) and \(OT_2 \perp \text{second line}\).

Step 4: This means the perpendicular distance from the centre \(O\) to the first line is equal to \(OT_1 = r\) (radius), and the perpendicular distance from \(O\) to the second line is also equal to \(OT_2 = r\).

Step 5: Therefore, the centre \(O\) is equidistant from the two lines.

Step 6: In geometry, the set (locus) of all points that are equidistant from two intersecting lines is the pair of their angle bisectors.

Step 7: Hence, the centre of the circle must lie on the angle bisector of the given two lines.

Step 1: Let the two common tangents \(AB\) and \(CD\) meet at a point \(P\) outside the circles.

Step 2: From geometry, we know that the lengths of tangents drawn from the same external point to a circle are equal.

Step 3: Apply this property to the left circle:

• \(PA\) and \(PC\) are tangents from point \(P\).
• Therefore, \(PA = PC\).

Step 4: Apply the same property to the right circle:

• \(PB\) and \(PD\) are tangents from point \(P\).
• Therefore, \(PB = PD\).

Step 5: Look at line \(PAB\):

• \(AB = PA - PB\).

Step 6: Now look at line \(PCD\):

• \(CD = PC - PD\).

Step 7: Since \(PA = PC\) and \(PB = PD\), subtracting gives:

\[ AB = PA - PB = PC - PD = CD \]

Final Result: Hence, we have proved that \(AB = CD\).

Step 1: We are given two circles of equal radius. Let the radius of each circle be \(r\) (in SI units, radius is measured in metres, \(m\)).

Step 2: Draw their common external tangents. These tangents will touch the first circle at points \(A\) and \(B\), and the second circle at points \(C\) and \(D\).

Step 3: Because the radii of the circles are equal, the distance of the centre of each circle from any tangent is the same, equal to \(r\).

Step 4: This means that the two external tangents are parallel lines. Why? Because each tangent is at the same distance from both centres.

Step 5: Now consider the two parallel tangents. The line joining the centres of the circles will cut both tangents at right angles. This line works like a transversal crossing two parallel lines.

Step 6: By symmetry, the distance between the points of contact on the first tangent (segment \(AB\)) is equal to the distance between the points of contact on the second tangent (segment \(CD\)).

Step 7: Therefore, we conclude that \(AB = CD\).

Note: This result can also be seen as a special case of Question 5 when the external centre of similarity goes to infinity.

Step 1: Recall a basic property of tangents: From a point outside a circle, the two tangents drawn to that circle are always equal in length.

Step 2: Here, point \(E\) is a common external point. - For the left circle: the two tangents are \(EA\) and \(EC\). So, \(EA = EC\).

Step 3: For the right circle: the two tangents from \(E\) are \(EB\) and \(ED\). So, \(EB = ED\).

Step 4: Now look at line segment \(AB\). It can be written as: \(AB = EA - EB\).

Step 5: Similarly, line segment \(CD\) can be written as: \(CD = EC - ED\).

Step 6: Since we already know that \(EA = EC\) and \(EB = ED\), we can replace them: \(AB = EA - EB = EC - ED = CD\).

Final Result: Therefore, \(AB = CD\).

Step 1: Draw a circle with center \(O\). Mark chord \(PQ\) and a tangent at point \(R\). Given: \(PQ \parallel \text{tangent at } R\).

Step 2: Since the chord \(PQ\) is parallel to the tangent at \(R\), the angle made at \(R\) by line \(RP\) (say \(\angle QRP\)) will be equal to the angle at \(P\) inside the triangle (\(\angle QPR\)). These are alternate interior angles from parallel lines.

Step 3: Now apply the Tangent–Chord Theorem: The angle between a tangent and a chord through the point of contact is equal to the angle in the opposite arc. So, \(\angle QRP = \angle PQR\).

Step 4: From Step 2, we already have \(\angle QRP = \angle QPR\). From Step 3, we have \(\angle QRP = \angle PQR\). Therefore, \(\angle QPR = \angle PQR\).

Step 5: This means triangle \(PQR\) is isosceles (two angles equal). So, arc \(PR\) = arc \(RQ\).

Final Step: Since both arcs are equal, point \(R\) divides the bigger arc \(PRQ\) into two equal parts. Hence, \(R\) bisects arc \(PRQ\).

Step 1: Draw a circle with center \(O\). Take a chord \(PQ\) of this circle.

Step 2: At point \(P\), draw a tangent to the circle. Similarly, at point \(Q\), draw a tangent to the circle.

Step 3: Let the angle made by the tangent at \(P\) with the chord \(PQ\) be \(\alpha\). Let the angle made by the tangent at \(Q\) with the chord \(PQ\) be \(\beta\).

Step 4: Recall the Tangent–Chord Theorem: "The angle between a tangent and a chord at the point of contact is equal to the angle in the alternate segment of the circle."

Step 5: Apply this theorem at point \(P\): The angle between the tangent at \(P\) and the chord \(PQ\) (i.e., \(\alpha\)) is equal to the angle formed at the opposite arc (say at point \(R\) on arc \(Q\widehat{P}\)).

Step 6: Apply the theorem at point \(Q\): The angle between the tangent at \(Q\) and chord \(PQ\) (i.e., \(\beta\)) is equal to the angle formed at the opposite arc (say at the same point \(R\) on arc \(P\widehat{Q}\)).

Step 7: Both \(\alpha\) and \(\beta\) are equal to the same angle at point \(R\). Therefore, \(\alpha = \beta\).

Final Result: The tangents drawn at the ends of a chord of a circle make equal angles with the chord.

Step 1: Draw a circle with center \(O\). Mark a point \(A\) on the circle.

Step 2: Draw the diameter \(AB\) of the circle, passing through \(A\) and the center \(O\).

Step 3: At point \(A\), draw a tangent to the circle. By definition, a tangent touches the circle at exactly one point.

Step 4: From geometry, the tangent at \(A\) is always perpendicular to the radius \(OA\).

Step 5: Now, draw a chord (say \(CD\)) of the circle such that it is parallel to the tangent at \(A\).

Step 6: Since \(CD\) is parallel to the tangent at \(A\), and the tangent is perpendicular to \(OA\), this means \(CD\) is also perpendicular to \(OA\).

Step 7: In circle geometry, a radius (or diameter) that is perpendicular to a chord always bisects the chord. That is, it divides the chord into two equal parts.

Step 8: Here, \(OA\) (or the diameter \(AB\)) is perpendicular to chord \(CD\). Therefore, diameter \(AB\) must pass through the midpoint of \(CD\).

Final Conclusion: Thus, the diameter \(AB\) bisects every chord of the circle that is parallel to the tangent at \(A\).