Circles: Solved Examples

Q.1. Two circles with same centre P have radii 6.5 cm and 3.3 cm. Through a point A of the larger circle, a tangent is drawn to the smaller circle touching it at B. Find AC.
A. 8.3 cm
B. 11.2 cm
C. 5.5 cm
D. 16.7 cm
Answer: Option B
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Sol :
Circle Concepts & Tricks
∠PBC = 90° (A tangent is perpendicular to the radius at the point of contact)
So (6.5)2 = (3.3)2 + (BC)2.
So BC = 5.6. Hence AC = 2 × 5.6 = 11.2 cm.
Q.2. Find the area of the circle having center at (5, 7) and passing through (2, 3).
A. 16π
B. 20π
C. 21π
D. 25π
Answer: Option D
Sol :
Centre at (5, 7) and passing through (2, 3). So, radius of circle is r =√(2-5)2 + (3-7)2 = 5. So, the area of circle = 25π.
Q.3. In a circle of radius 10, A & B are two points on the circumference. If arc AB is 120°, what is the length of line segment AB?
A. 10√3;
B. 10
C. 5√3;
D. None of these
Answer: Option C
Sol :
Half of the triangle formed between A, B and the centre of the circle will be a 30 – 60 – 90 triangles. Since 10 (the radius) is the hypotenuse of this triangle, half of AB will be 10 × √3 / 2 = 5√3.
Q.4. AB is a chord of the circle with centre O such that the length of the minor arc AB is 2π metres. If ∠AOB = 90°, what is the area of the minor sector AOB?
A. 24πm2
B. 4πm2
C. 16πm2
D. 6πm2
Answer: Option B
Sol :
Since the central angle is 90° and the length of the corresponding arc is 2π, the circumference of the circle is 8π. So the radius of the circle is 4 m. The area of the sector is therefore ¼ × π × 42 = 4π m2.
Q.5. AP is tangent at point P to the circle with centre O and diameter BC. If AP = 20 and AB = 10, what is the length of AO?
Circle Concepts & Tricks
A. 25
B. 15
C. 35
D. 20
Answer: Option A
Sol :
Suppose OP = r. Then, OA = 10 + r.
In ΔAPO, (10 + r)2 = 400 + r2.
Solving this equation gives r = 15. So, AO = 10 + 15 = 25.
Q.6. In the figure, P, Q and R are the centres of 3 circles such that P, Q and R are collinear. What is the ratio of the area of the entire shaded region to the area of the unshaded region?
Circle Concepts & Tricks
A. 5:2
B. 17:4
C. 13:3
D. 17:3
Answer: Option C
Sol :
If the radius of circle R is r, then the radii of circles P and Q are 4r and 2r respectively. The unshaded region is 4πr2 – πr2 = 3πr2. The entire shaded region is 16πr2 – 3πr2 = 13πr2. The ratio is 13: 3.
Q.7. A chord AB of a circle of radius 10.5 cm makes an angle of 120° at the centre of the circle. Find the area of the minor segment approximately.
A. 62.5 cm2
B. 60.25 cm2
C. 67.7 cm2
D. 61.5 cm2
Answer: Option C
Sol :
As the central angle is given to be 120°, you must quickly realize that it will result into an equilateral triangle.
Now area of the minor sector: π × (10.5)2 &time' 120 / 360 = 115.4
Area of equilateral triangle = (10.5)2 × √3/4 = 47.7
The difference of the two represents the area of the minor segment (Note: it is not a sector)
i.e. 115.4 – 47.7 = 67.7 cm2
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Q.8. Find the approximate area of the major segment as per the information given in the above question?
A. 278.5 cm2
B. 242.25 cm2
C. 262 cm2
D. 247 cm2
Answer: Option A
Sol :
The area of the major segment
= Area of the circle – area of the minor segment → × × (10.5)2 = 346.2.
Area of major segment = 346.2 – 67.7 = 278.5 cm2.
Q.9. It is not possible to draw a circle having its centre on a fixed straight line M and passing through two fixed points P and Q not on M if
A. M is parallel to PQ
B. M is the perpendicular bisector of PQ
C. M is perpendicular to PQ but does not bisect it
D. M is not perpendicular to PQ but bisects it
Answer: Option C
Sol :
In all the remaining cases, except (C), we can draw at least one circle that is on M and passes through P and Q. In case M is perpendicular to PQ, but does not bisect it, then we will have no point on M that is equidistant from both P and Q.
Q.10. Two small circular parks of diameters 32 m and 24 m are to be replaced by a bigger circular park. What would be the radius of this new park, if the new park has the same area as the two small parks?
A. 5
B. 10
C. 15
D. 20
Answer: Option D
Sol :
Area of the new circular park = Sum of the areas of the 2 smaller parks ⇒ π (32/2)2 + π (24/2)2 = π(256 + 144) = 400 π ⇒ 400 π = πR2. ∴ R2 = 400 ⇒ R = 20m.
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