CBSE Class 10 Circles Sure Shot Questions Set 05

Question. Two concentric circles have radii 5 cm and 3 cm. Find the length of the chord of the larger circle which touches the smaller (circle at a point).
(a) 4
(b) 5
(c) 8
(d) 10

Answer: (c) 8
Explanation : Given, \( OA = OB = 5 \) cm; \( OD = 3 \) cm. AB is a tangent at D, smaller circle and chord for bigger circle. \( \therefore \angle ODB = \angle ODA = 90^\circ \) [As the tangent at a point is perpendicular to the radius of a circle through the point]. By pythagoras theorem \( OB^2 = BD^2 + OD^2 \Rightarrow BD^2 = OB^2 - OD^2 \Rightarrow BD^2 = (5)^2 - (3)^2 \Rightarrow BD^2 = 25 - 9 = 16 \Rightarrow BD = 4 \) cm. Also, a radius from the centre bisects the chord. \( \therefore AB = AD + BD = 2BD = (2 \times 4) \) cm \( = 8 \) cm.

Question. A chord of a circle of radius 10 cm subtends a right angle at its centre. The length of the chord (in cm) is :
(a) \( 5\sqrt{2} \)
(b) \( 10\sqrt{2} \)
(c) \( \frac{5}{\sqrt{2}} \)
(d) \( 10\sqrt{3} \)

Answer: (b) \( 10\sqrt{2} \)
Explanation : Given, \( AO = OB = 10 \) cm and \( \angle AOB = 90^\circ \). Thus, \( AB^2 = AO^2 + OB^2 \) [By Pythagoras Theorem]. \( \Rightarrow AB^2 = (10)^2 + (10)^2 \Rightarrow AB^2 = 100 + 100 \Rightarrow AB^2 = 200 \). Thus \( AB = \sqrt{200} = 10\sqrt{2} \) cm.

Question. Two circles touch each other externally at P. AB is a common tangent to the circles, touching them at A and B. The value of \( \angle APB \) is :
(a) 30°
(b) 45°
(c) 60°
(d) 90°

Answer: (d) 90°
Explanation : Let PO meet AB at O such that PO is a tangent. Thus, \( AO = PO = OB \) and \( \angle AOP = \angle BOP = 90^\circ \). As \( \Delta AOP \) and \( \Delta BOP \) are right-angled isosceles triangles. So, \( \angle OAP = \angle OPA = \angle OBP = \angle OPB = 45^\circ \). Thus, \( \angle APB = \angle APO + \angle OPB = 45^\circ + 45^\circ = 90^\circ \).

Question. From a point Q, 13 cm away from the centre of a circle, the length of tangent PQ to the circle is 12 cm. The radius of the circle (in cm) is :
(a) 25
(b) \( \sqrt{313} \)
(c) 5
(d) 1

Answer: (c) 5
Explanation : Given, \( OQ = 13 \) cm and \( PQ = 12 \) cm. In \( \Delta OPQ \), by pythagoras theorem, radius \( r = OP = \sqrt{OQ^2 - PQ^2} = \sqrt{(13)^2 - (12)^2} = \sqrt{169 - 144} = \sqrt{25} = 5 \) cm. As length cannot be negative.

Question. If radii of two concentric circles are 4 cm and 5 cm, then the length of each chord of one circle which is tangent to the other circle is :
(a) 3 cm
(b) 6 cm
(c) 9 cm
(d) 1 cm
Answer: (b) 6 cm
Explanation :
Let O be the centre of two concentric circles \(C_1\) and \(C_2\) whose radii are \(r_1 = 4 \text{ cm}\) and \(r_2 = 5 \text{ cm}\), Now, we draw a chord AC of circle \(C_2\), which touches the circles at B. Also, join OB, which is perpendicular to AC. [Tangent at any point of circle is perpendicular to radius throughly the point of contact]
Now, in right angled OBC, by using Pythagoras theorem,
\(\therefore OC^2 = BC^2 + BO^2\)
[\(\because (\text{hypotenuse})^2 = (\text{base})^2 + (\text{perpendicular})^2\)]
\(\Rightarrow 5^2 = BC^2 + 4^2\)
\(\Rightarrow BC^2 = 25 - 16 = 9\)
\(\Rightarrow BC = 3 \text{ cm}\)
\(\therefore \text{Length of chord } AC = BC = 2 \times 3 = 6 \text{ cm}\)

Question. From a point P which is at a distance of 13 cm from the centre O of a circle of radius 5 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) 60 cm²
Explanation :
Firstly, draw a circle of radius 5 cm having centre OP is a point distance of 13 cm from O. A pair of tangents PQ and PR are drawn. Thus, quadrilateral PQOR is formed.
\(\because OQ \perp QP\) [since, AP is a tangent line]
In right angled \(\triangle PQO, OP^2 = OQ^2 + PQ^2\)
\(\Rightarrow 13^2 = 5^2 + PQ^2\)
\(\Rightarrow PQ^2 = 169 - 25 = 144\)
\(\Rightarrow PQ = 12 \text{ cm}\)
Now, area of \(\triangle OQP = \frac{1}{2} \times QP \times QO\)
\(= \frac{1}{2} \times 12 \times 5 = 30 \text{ cm}^2\)
\(\therefore \text{Area of quadrilateral } PQOR = 2 \times \triangle OQP\)
\(= 2 \times 30 = 60 \text{ cm}^2\)

Question. The tangent of a circle makes-angle with radius at point of contact :
(a) 45°
(b) 30°
(c) 90°
(d) 60°
Answer: (c) 90°
Explanation :
The angle between a tangent to a circle and the radius through the point of contact is always a right angle or 90°.

Question. The angle subtended by the diameter of a semi-circle is :
(a) 90°
(b) 45°
(c) 180°
(d) 60°
Answer: (c) 180°
Explanation :
The semicircle is half of the circle; hence the diameter of the semicircle will be a straight-line subtending 180 degrees.

Question. If chords AB and CD of congruent circles subtend equal angles at their centres, then :
(a) AB = CD
(b) AB > CD
(c) AB < AD
(d) None of the options
Answer: (a) AB = CD
Explanation :
Take the reference of the figure from above question.
In \(\triangle AOB\) and \(\triangle COD\),
\(\angle AOB = \angle COD\) (given)
\(OA = OC\) and \(OB = OD\) (Radii of the circle)
So, \(\triangle AOB \cong \triangle COD\). (SAS congruency)
\(\therefore AB = CD\) (By CPCT)

Question. An equilateral triangle of side 9 cm is inscribed in a circle. The radius of the circle is :
(a) 3 cm
(b) \(3\sqrt{2} \text{ cm}\)
(c) \(3\sqrt{3} \text{ cm}\)
(d) 6 cm
Answer: (c) \(3\sqrt{3} \text{ cm}\)
Explanation :
Let \(\triangle ABC\) be an equilateral triangle.
Let AD be one of its medians.
Then, \(AD \perp BC\) and \(BD = 4.5 \text{ cm}\)
In right \(\triangle ADB\),
\(\therefore AD = \sqrt{AB^2 - BD^2}\) ....(By Pythagoras theorem)
AD = \(\sqrt{9^2 - 4.5^2} = \sqrt{81 - 20.25} = \sqrt{60.75} = 4.5\sqrt{3} \text{ cm}\).
The radius of the circumcircle (G being the centroid) is \(AG = \frac{2}{3}AD = \frac{2}{3} \times 4.5\sqrt{3} = 3\sqrt{3} \text{ cm}\).

Question. If tangents PA and PB from a point P to a circle with centre O are inclined to each other at angle of 80°, then \( \angle POA \) is equal to :
(a) 50°
(b) 60°
(c) 70°
(d) 80°
Answer: (a) 50°
Explanation :
First, draw the diagram according to the given statement.
Now, in the above diagram, OA is the radius to tangent PA and OB is the radius to tangent PB.
So, OA is perpendicular to PA and OB is perpendicular to PB i.e., \( OA \perp PA \) and \( OB \perp PB \)
So, \( \angle OBP = \angle OAP = 90^\circ \)
Now, in the quadrilateral AOBP,
\( \angle AOB + \angle OAP + \angle OBP + \angle APB = 360^\circ \)
\( \therefore \angle AOB + 260^\circ = 360^\circ \)
\( \Rightarrow \angle AOB = 100^\circ \)
Now, consider the triangles \( \triangle OPB \) and \( \triangle OPA \). Here,
AP = BP
(Since, the tangents from a point are always equal)
OA = OB
(Which are the radii of the circle)
OP = OP
(It is the common side)
\( \therefore \triangle OPB \cong \triangle OPA \)
(by SSS congruency)
So, \( \angle POB = \angle POA \)
Now, \( \angle AOB = \angle POA + \angle POB \)
\( 2(\angle POA) = \angle AOB \) [from (i)]
By putting the respective values, we get,
\( \Rightarrow \angle POA = 100^\circ/2 = 50^\circ \)

Question. A tangent PQ at a point P of a circle of radius 5 cm meets a line through the centre O at a point Q so that OQ = 12 cm. Lenght PQ is :
(a) 12 cm
(b) 13 cm
(c) 8.5 cm
(d) \( \sqrt{119} \) cm
Answer: (d) \( \sqrt{119} \) cm
Explanation :
In the above figure, the line that is drawn from the centre of the given circle to the tangent PQ is perpendicular to PQ.