A. 8.3 cm

B. 11.2 cm

C. 5.5 cm

D. 16.7 cm

Sol :

∠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.

∠PBC = 90° (A tangent is perpendicular to the radius at the point of contact)

So (6.5)

So BC = 5.6. Hence AC = 2 × 5.6 = 11.2 cm.

A. 16π

B. 20π

C. 21π

D. 25π

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.

Centre at (5, 7) and passing through (2, 3). So, radius of circle is r =√(2-5)

A. 10√3;

B. 10

C. 5√3;

D. None of these

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.

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.

A. 24πm^{2}

B. 4πm^{2}

C. 16πm^{2}

D. 6πm^{2}

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 × 4^{2} = 4π m^{2}.

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 × 4

A. 25

B. 15

C. 35

D. 20

Sol :

Suppose OP = r. Then, OA = 10 + r.

In ΔAPO, (10 + r)^{2} = 400 + r^{2}.

Solving this equation gives r = 15. So, AO = 10 + 15 = 25.

Suppose OP = r. Then, OA = 10 + r.

In ΔAPO, (10 + r)

Solving this equation gives r = 15. So, AO = 10 + 15 = 25.

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A. 5:2

B. 17:4

C. 13:3

D. 17:3

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 4πr^{2} – πr^{2} = 3πr^{2}. The entire shaded region is 16πr^{2} – 3πr^{2} = 13πr^{2}. The ratio is 13: 3.

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 4πr

A. 62.5 cm^{2}

B. 60.25 cm^{2}

C. 67.7 cm^{2}

D. 61.5 cm^{2}

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 cm^{2}

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)

Area of equilateral triangle = (10.5)

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 cm

A. 278.5 cm^{2}

B. 242.25 cm^{2}

C. 262 cm^{2}

D. 247 cm^{2}

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 cm^{2}.

The area of the major segment

= Area of the circle – area of the minor segment → × × (10.5)

Area of major segment = 346.2 – 67.7 = 278.5 cm

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

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.

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.

A. 5

B. 10

C. 15

D. 20

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 = πR^{2}. ∴ R^{2} = 400 ⇒ R = 20m.

Area of the new circular park = Sum of the areas of the 2 smaller parks ⇒ π (32/2)