CBSE Class 10 Mathematics Circles VBQs Set C

Multiple Choice Questions

Question. From an external point Q, the length of the tangents to a circle is \( 5 \text{ cm} \) and the distance of Q from the centre is \( 8 \text{ cm} \). The radius of the circle is:
(a) \( 39 \text{ cm} \)
(b) \( \sqrt{3} \text{ cm} \)
(c) \( \sqrt{39} \text{ cm} \)
(d) \( 7 \text{ cm} \)
Answer: (c)
Explanation:
Here \( QA = 5 \text{ cm} \), \( OQ = 8 \text{ cm} \).
since, \( \angle OAQ = 90^\circ \)
\(\therefore\) In right \( \Delta OAQ \), we have
\( OA^2 = OQ^2 - QA^2 = 8^2 - 5^2 = 64 - 25 = 39 \)
\(\therefore\) Radius of the circle = \( OA = \sqrt{39} \text{ cm} \).

Question. From a point P which is at a distance of \( 13 \text{ cm} \) from the centre O of a circle of radius \( 5 \text{ cm} \), the pair of tangents PQ and PR to the circle are drawn. Then the area of the quadrilateral PQOR is:
(a) \( 60 \text{ cm}^2 \)
(b) \( 65 \text{ cm}^2 \)
(c) \( 30 \text{ cm}^2 \)
(d) \( 32.5 \text{ cm}^2 \)
Answer: (a)
Explanation: We know that tangent to at any point on the circle is perpendicular to the radius through the point of contact. Hence, we get \( OQ \perp PQ \) and \( OR \perp PR \).
\( \Delta POQ \) and \( \Delta POR \) are right-angled triangles.
Using Pythagoras theorem in \( \Delta PQO \):
\( (\text{Base})^2 + (\text{Perpendicular})^2 = (\text{Hypotenuse})^2 \)
\( (PQ)^2 + (OQ)^2 = (OP)^2 \)
\( (PQ)^2 + (5)^2 = (13)^2 \)
\( (PQ)^2 + 25 = 169 \Rightarrow (PQ)^2 = 144 \Rightarrow PQ = 12 \text{ cm} \).
We know that tangents through an external point to a circle are equal, \( PQ = PR = 12 \text{ cm} \).
Therefore, area of quadrilateral PQRS, \( A = \text{area of } \Delta POQ + \text{area of } \Delta POR \).
Area of right angled triangle \( = \frac{1}{2} \times \text{base} \times \text{perpendicular} \)
\( A = (\frac{1}{2} \times OQ \times PQ) + (\frac{1}{2} \times OR \times PR) \)
\( A = (\frac{1}{2} \times 5 \times 12) + (\frac{1}{2} \times 5 \times 12) \)
\( A = 30 + 30 = 60 \text{ cm}^2 \)

Question. The chord of a circle of radius \( 10 \text{ cm} \) subtends a right angle at its centre. The length of the chord (in cm) is?
(a) \( 10 \text{ cm} \)
(b) \( 10\sqrt{2} \text{ cm} \)
(c) \( 20 \text{ cm} \)
(d) \( 12 \text{ cm} \)
Answer: (b)
Explanation:
In \( \Delta POQ \), By Pythagoras theorem,
\( PQ^2 = PO^2 + OQ^2 \)
\( PQ^2 = 10^2 + 10^2 \)
\( PQ^2 = 200 \)
\( PQ = 10\sqrt{2} \text{ cm} \)
So, the length of the chord is \( 10\sqrt{2} \text{ cm} \).

Question. At one end A of diameter AB of a circle of radius \( 5 \text{ cm} \), tangent XAY is drawn to the circle. The length of the chord CD parallel to XY and at a distance \( 8 \text{ cm} \) from A is:
(a) \( 4 \text{ cm} \)
(b) \( 5 \text{ cm} \)
(c) \( 6 \text{ cm} \)
(d) \( 8 \text{ cm} \)
Answer: (d)
Explanation: According to the question, Radius of circle, \( AO = OC = 5 \text{ cm} \). \( AE = 8 \text{ cm} \).
\( AE = AO + OE \Rightarrow OE = AE - AO = 8 - 5 = 3 \text{ cm} \).
Now, \( \angle OAX = \angle BAX = 90^\circ \) [Tangent at any point of a circle is perpendicular to the radius through the point of contact].
Also, \( CD \parallel XY \) meets AB at E, so we have \( 90^\circ + \angle AEC = 180^\circ \Rightarrow \angle AEC = 90^\circ \Rightarrow \angle OEC = 90^\circ \).
By Pythagoras theorem, \( OC^2 = OE^2 + EC^2 \Rightarrow (5)^2 = (3)^2 + (EC)^2 \Rightarrow EC^2 = 25 - 9 = 16 \Rightarrow EC = 4 \).
Also, \( CE = ED \) [since, perpendicular from center to the chord bisects the chord].
\( CD = 2 \times EC = 8 \text{ cm} \).

Question. If radii of concentric circles are \( 4 \text{ cm} \) and \( 5 \text{ cm} \), then find the length of each chord of one circle which is tangent to the other circle.
(a) \( 8 \text{ cm} \)
(b) \( 6 \text{ cm} \)
(c) \( 10 \text{ cm} \)
(d) \( 12 \text{ cm} \)
Answer: (b)
Explanation: \( OA = 4 \text{ cm} \), \( OB = 5 \text{ cm} \). Also, \( OA \perp BC \).
\(\therefore OB^2 = OA^2 + AB^2 \Rightarrow 5^2 = 4^2 + AB^2 \Rightarrow AB = \sqrt{25 - 16} = 3 \text{ cm} \).
\( BC = 2 AB = 2 \times 3 = 6 \text{ cm} \).

Fill in the Blanks

Question. All concentric circles are ....................... to each other. 
Answer: Similar
Explanation: All concentric circles are similar.

Question. In the figure, \( \Delta ABC \) is circumscribing a circle, the length of BC is ....................... cm. 
Answer: 10
Explanation: Here, \( BC = BQ + QC = BP + CR \) as tangents from an external point are equal. (\(\because BQ = BP, QC = CR\)).
\(\therefore BC = 3 + (11 - AR) \) (\(\because CR = AC - AR\))
\( BC = 14 - AR \)
\( BC = 14 - AP \) (\(\because AR = AP\))
\( BC = 14 - 4 = 10 \)

Question. ABCD is a cyclic quadrilateral. If \( \angle BAC = 50^\circ \) and \( \angle DBC = 60^\circ \) then \( \angle BCD = ....................... . \)
Answer: \( 70^\circ \)
Explanation: Here \( \angle BDC = \angle BAC = 50^\circ \) (angles in same segment are equal). In \( \Delta BCD \), we have \( \angle BCD = 180^\circ - (\angle BDC + \angle DBC) = 180^\circ - (50^\circ + 60^\circ) = 180^\circ - 110^\circ = 70^\circ \).

Question. The length of the tangent to a circle from a point P, which is \( 25 \text{ cm} \) away from the centre, is \( 24 \text{ cm} \). The radius of the circle is ........................
Answer: \( 7 \text{ cm} \)
Explanation: \(\because OQ\) is perpendicular to PQ. \( PQ^2 + OQ^2 = OP^2 \Rightarrow 24^2 + OQ^2 = 25^2 \Rightarrow OQ^2 = 625 - 576 = 49 \Rightarrow OQ = 7 \text{ cm} \).

Question. A tangent PQ at a point P of a circle of radius \( 5 \text{ cm} \) meets a line through the centre O at a point Q so that \( OQ = 13 \text{ cm} \). Then, the length PQ is ....................... .
Answer: \( 12 \text{ cm} \)
Explanation: Given, \( OP = 5 \text{ cm} \) [radius], \( OQ = 13 \text{ cm} \). Now, In \( \Delta OPQ, \angle P = 90^\circ \) [Radius is perpendicular to tangent at the point of contact]. By pythagoras theorem, \( (OQ)^2 = (OP)^2 + (PQ)^2 \Rightarrow PQ = \sqrt{13^2 - 5^2} = \sqrt{169 - 25} = \sqrt{144} = 12 \text{ cm} \).

Question. Secant intersects a circle at ....................... distinct points.
Answer: Secant intersects a circle at two distinct points.

Write True or False

Question. If a chord AB subtends an angle of \( 60^\circ \) at the centre of a circle, then the angle between the tangents at A and B is also \( 60^\circ \).
Answer: False.
Explanation: We have a circle with center O and AB as any chord with \( \angle AOB = 60^\circ \). Now, \( OA \perp AC \) and \( OB \perp BC \) [As tangent at any point on the circle is perpendicular to the radius through point of contact]. \( \angle OBC = \angle OAC = 90^\circ \) ...(i). In quadrilateral AOBC, \( \angle OBC + \angle OAC + \angle AOB + \angle ACB = 360^\circ \) [By angle sum property of quadrilateral]. \( 90^\circ + 90^\circ + 60^\circ + \angle ACB = 360^\circ \Rightarrow \angle ACB = 120^\circ \) ...(ii). i.e., the angle between two tangents is \( 120^\circ \). Hence, the given statement is false.

Question. The length of tangent from an external point on a circle is always greater than radius of the circle.
Answer: False.
Explanation: Consider any point P external to a circle away from O. Now, draw tangent PT on the circle. Clearly, the length of the tangent from an external point of a circle may or may not be greater than the radius of the circle. Hence, the given statement is false.

Question. The length of the tangent from an external point P on a circle with centre O is always less than OP. 
Answer: True.
Explanation: Consider the circle with centre O. Let PT be a tangent drawn from an external point P. Join OT. \( OT \perp PT \) [Tangent at any point on the circle is perpendicular to the radius through point of contact]. \( \Delta OPT \) is a right-angled triangle. We also know that in a right angled triangle, hypotenuse is always greater than any of the two sides of the triangle. \( \therefore OP > PT \Rightarrow PT < OP \). Hence, the length of tangent from an external point P on a circle with center O is always less than OP.

Question. The angle between two tangents to a circle may be \( 0^\circ \). 
Answer: True.
Explanation: The angle between two tangents to a circle may be \( 0^\circ \) only when both tangent lines coincide or are parallel to each other. Consider the diameter POQ of a circle with centre O. The tangent at P and Q are drawn. \( \Rightarrow OP \perp AB \) and \( OQ \perp CD \) [Tangent at any point on the circle is perpendicular to the radius through point of contact]. \( \angle OPA = 90^\circ \) and \( \angle OQD = 90^\circ \Rightarrow \angle OPA = \angle OQD = 90^\circ \). These are alternate interior angles, so the tangent AB \( \parallel \) CD i.e., angle between two tangents to a circle may be zero. Hence, the given statement is true.

Question. The tangent to the circumcircle of an isosceles \( \Delta ABC \) at A, in which AB = AC, is parallel to BC. 
Answer: True.
Explanation: Let us consider a circle in which EF is a tangent passing through point A on the circle and ABC is an isosceles triangle in the circle, in which AB = AC. \( \angle ACB = \angle ABC \) [Angles opposite to equal sides are equal]....(i). \( \angle EAB = \angle ACB \) [Angle between tangent and chord is equal to angle made by chord in alternate segment]. But \( \angle ABC = \angle ACB \Rightarrow \angle EAB = \angle ABC \Rightarrow EAF \parallel BC \) [Two lines are parallel if their alternate interior angles are equal]. Hence, the given statement is true.

Question. If a number of circles pass through the end points P and Q of a line segment PQ, then their centres lie on the perpendicular bisector of PQ. 
Answer: True.
Explanation: Centre of any circle passing through the end points P and Q of a line segment are equidistant from P and Q. We draw two circles with centre \( C_1 \) and \( C_2 \) passing through the end points P and Q of a line segment PQ. We know that perpendicular bisectors of a chord of a circle always passes through the centre of circle. Thus, perpendicular bisector of PQ passes through \( C_1 \) and \( C_2 \). i.e., \( C_1P = C_1Q, C_2P = C_2Q \). Similarly, all the circles passing through PQ will have their centre on perpendicular bisector of PQ.

Question. AB is the diameter of a circle and AC is its chord such that \( \angle BAC = 30^\circ \). If the tangent at C intersects AB extended at D, then BC = BD.
Answer: True.
Explanation: AB is the diameter of circle with centre O and AC is a chord such that \( \angle BAC = 30^\circ \). Also, tangent at C intersects AB extended at D. Join OC. \( OA = OC \) [Radii of the same circle]. \( \angle OCA = \angle OAC = 30^\circ \) [Angles opposite to equal sides are equal]. \( \angle ACB = 90^\circ \) [Angle in a semicircle is a right angle]. \( \angle ABC + \angle ACB + \angle CAB = 180^\circ \) [Angle sum property]. \( \angle ABC + 90^\circ + 30^\circ = 180^\circ \Rightarrow \angle ABC = 180^\circ - 120^\circ = 60^\circ \). \( \angle OCA + \angle OCB = 90^\circ \Rightarrow 30^\circ + \angle OCB = 90^\circ \Rightarrow \angle OCB = 60^\circ \). \( OC = OB \) [Radii of same circle]. \( \angle OBC = \angle OCB = 60^\circ \) [Angles opposite to equal sides are equal]. \( \angle OBC + \angle CBD = 180^\circ \) [Linear pair]. \( 60^\circ + \angle CBD = 180^\circ \Rightarrow \angle CBD = 120^\circ \). \( OC \perp CD \) [Tangent at a point on the circle is perpendicular to the radius through point of contact]. \( \angle OCD = 90^\circ \). \( \angle OCB + \angle BCD = 90^\circ \Rightarrow 60^\circ + \angle BCD = 90^\circ \Rightarrow \angle BCD = 30^\circ \). In \( \Delta BCD \), \( \angle CBD + \angle BCD + \angle BDC = 180^\circ \) [Angle sum property of triangle]. \( 120^\circ + 30^\circ + \angle BDC = 180^\circ \Rightarrow \angle BDC = 30^\circ \). From eqn. (iii) and eqn. (iv) \( \angle BCD = \angle BDC = 30^\circ \Rightarrow BC = BD \) [Sides opposite to equal angles are equal]. Hence, the given statement is true.