Circles

CONCEPTS TO BE COVERED:-

Introduction, Tangent to a circle, Number of Tangents from a point on a circle, secant to a circle, Number of secants, circumference, chord, Radius, diameter, Major& Minor segments, Sectors and Arc and MCQ's

INTRODUCTION:-

Circles is a collection of all points in a plane. which are at a same distance from a fixed point

we can draw number of lines to a circles which will intersect at different points. lets see some examples:

In the i, case the line doesn't intersect any point on the circle i.e Non - intersecting
In the ii, case there are two common points A and B on the line PQ which the circle
In the iii, case there is only one point on the line PQ which intersects the circle

Tangent to a circle:- It is a line that intersects the circle at only one point. The point where tangent touches the circles is called the point of contant.

Here A is the point of contact.

'PB' is the tangent to the circle

*Number of Tangents:- Infinitely many tangents can be drawn on a circle

*Secant to a circle:- A line which intersects a circle in two distinct points is called a secant of the circle

*Number of secant:- There are infinitely many secants which can be drawn on a circle.

*Circumference of circle:- The circumference of a circle is the perimeter of the circle it is the total length of the boundary of the circle

The formula for finding circumference of circle is 2πr

Example:- if the radius of a circle is 28cm find the circumference of the circle.

sol:-Given Radius of the circle = 28cm

circumference formula = 2πr

2*22/7*28

=176cm

*Chord:-The line segment joining any two points on the circumference of the circle.

Note:- Diameter is the longest chord of a circle

*Radius:- Radius of a circle or sphere is any of the line segments from its center to its perimeter is called the Radius of the center

OA is the Radius of the circle

*Diameter:- Diameter is a line that passes passes through the center and meets the circumference at opposite. it is twice as long as that of the radius.

Diameter = 2* measurement of the radius

*Major segment:- The portion (or) part of the circular region enclosed between a chord and the corresponding Major arc is called the Major Segment of the circle.

*Minor segment:- The portion (or part) of the circular region enclosed between a chord and the corresponding minor arc is called as minor segment of the circle.

*Sector:- it is a part of a circle made of the arc of the circle along with its two radii we can compare the shape the shape of the sector with the piece of pizza

*Arc:- An arc is one of the portions of a circles . it is basically a part of the circumference of a circle .Arc is a part of a curve.

*The proofs of the following theorem can be asked in the examination:-

(a) The tangent at any point of a circle is perpendicular to the radius through the point of contact .

(b) The lengths of tangents drawn from an external points to a circle are equal.

*Theorem-1:-

The Tangent at any point of a circle is perpendicular to the radius through the point of contact

Proof:-

O is the centre of the circle of the circle and radius of the circle is fixed.

OC = OA (radius)

also OB = OC+ BC

so OC < OB

OA < OB ( since OA = OC )

Hence OA perpendicular I

Hence, the tangent at any point of a circle is perpendicular to the radius through the point of contact

*Sample problems based on the Theorem:-

problem 1: Find the angle CBA given that CA is a line drawn from the centre to the tangent on the circle . The length of the radius and the base length are mentioned in the question

sol:- Here we apply trigonometric formulas to get the Angle CBA

tan(Angle CBA) = CA / AB

tan(Angle CBA) = 3/4

Angle CBA = 37 degree

problem 2 : In the figure given , O is the center of the circle . from point R outside the circle, as shown RM and RN tangent touching the circle at M and N . if the length of OR = 10cm and radius of the circle = 5cm . then what is the length of each tangent ?

sol:- Given that OM= 5 cm and OR = 10cm

In right ∆OMR

OR^2 = OM^2 + MR^2

MR^=OR^2 - OM^2

MR^2 = 75

MR = 5√3 cm then NR = 5√3 cm

Hence the length of each tangent is 5√3 cm

*Theorem 2:-

Prove that the lengths of tangents drawn from an external point to a circle are equal .

Proof:-

Given that PT and TQ are two tangents drawn from an external point T to the circle

To prove : PT = TQ

we know that a tangent to the circle is perpendicular to the radius through the point of contact so, Traingle OPT = Triangle OQT

OT = OT (common)

Triangle OPT = Triangle OQT = 90 degrees ( tangent and radius are perpendicular at point of contact )

OP = OQ = radius

▵OPT = ▵OQT (RHS congruence)

PT = TQ ( by c.p.c.t)

so, length of the tangents drawn from an external point to circle are equal .

*Sample problems based on the theorem:-

problem:- In fig PA and PB are two tangents drawn from an external point p to a circle with Centre c and radius 4cm . if PA perpendicular PB , then the length of each tangent is:

sol:- Given that PA and PB are two tangent drawn from an external point p. And PA perpendicular to PB

join center c with A and B

given PA and PB are tangent of circle

Then PA perpendicular CA and PB perpendicular to CB

Therfore , Angle CAP = 90 degree and angle CBP = 90 degree

but given Angle APB = 90 degree

Therefore, Angle CBA = 90degree

therefore, quadrilateral ABCD has all

angles equal i.e 90 degree

and adjacent sides equal

hence it is a square

Therefore, length of each tangent will be 4cm

problem-2: If the angle between two tangents drawn from an external point p to a circle of radius 'a' and center o, is 60degree then find the length of OP .

sol:- Angle QPR = 90 degree

and Angle OPQ = 30 degree

sin30degree = OQ/OP

OP=OQ/Sin 30 = a / 1/2

= 2a

OP = 2a units

problem-3:- In figure PA and PB are two tangents drawn from an external point p to a circle with center c and radius 4 cm . if PA perpendicular to PB, then find the length of each tangent.

sol:- CA is perpendicular to AP and CB is perpendicular to BP

Again AC = BC= 4 (radius of the circle )

also AP=PB= (Tangents from point p )

so, BAPC is a square

AP=PB=BC=CA=4cm

so, length of tangents are 4cm each.

Some Example problems:-

Example:-1 Prove that in two concentric circles, the chord of the larger circle, which touches the smaller circle, is bisected at the point of contact.

Sol:- Given that two concentric circles c₁ and c₂ with Centre O and a chord AB of the larger circle c₂.

It touches the smaller circle c at the point P we need to rove that AP = BP. Join OP.

Then AB is a tangent to c₂ at p and OP is its radius.

OP perpendicular AB [The tangent at any point of a circle is perpendicular to the radius through the point of contact.]

AB is chord and OP is the bisector

(i.e.) AP=BP

Example 2:- Two tangents TP and TQ are drawn to a circle with Centre O from external point T. Prove that ∠ PTQ = 2∠ OPQ

Sol:- Given the circle with center O an External point and two tangents TP and TQ to the circle, where P, Q are the points of the contact. we need to prove that,

∠ PTQ = 2∠ OPQ
∠ PTQ = θ

By the theorem, the lengths of the tangents drawn from an external point to a circle are equal.

TP = TQ

so, TPQ is an isosceles triangle.

Therefore, ∠ TPQ = ∠ TQP = 1/2 (180⁰ - θ ) = 90⁰ - 1/2 θ

Also by the theorem,

The tangent at the any point of a circle is perpendicular to the radius through the point of contact.

∠ OPQ = ∠ OPT - ∠ TPQ = [90⁰ - 1/2 θ]

1/2 θ = 1/2 ∠ PTQ

∠ PTQ = 2∠ OPQ

Multiple choice questions [MCQ'S]

1.The radii of two circles are 19 cm and 9cm on respectively. The radius of the circle which has circumference equal to the sum of the circumference of two circles is

(a) 35cm

(b) 10cm

(d) 28 cm

2. The area of the circle that can be inscribed in a square of side 6cm, is

(a) 18π cm²

(b) 12π cm²

(c) 9π cm²

(d) 14π cm²

3. The area of the Circle is 154 cm². The radius of the circle is

(a) 7 cm

(b) 14 cm

(d) 17.5 cm

4. The area of the circle whose diameter is 21cm is

(a) 346.5 cm²

(b) 37.68 cm²

(d) 19.84 cm²

5. The perimeter of a Circle having radius 5cm is equal to:

(a) 30 cm

(b) 3.14 cm

(d) 40 cm

6. Area of the circle with radius 5 cm is equal to:

(a) 60 sq.cm

(b) 75.5 sq.cm

(c) 78.5 sq.cm

(d) 10.5 sq.cm

7. The area of the circle that can be inscribed in a square of side 8cm is

(a) 36π cm²

(b) 16π cm²

c) 12π cm²

(d) 9π cm²

8. The area of the square that can be inscribed in a circle of radius 8 cm is

(a) 256 cm²

(b) 128 cm²

(d) 64 cm²

9. If the sum of area of two circles with radii R₁ and R₂ is equal to the area of circle of radius R₁ then

(a) R₁ + R₂ = R²

(b) R₁² + R₂² = R²

(d) R₁² + R₂² < R²

10. If the circumference of a circle and the perimeter of a square are equal, then

(a) area of the of the square = Area of the square

(b) area of circle > area of square

(d) Nothing definite can be said about the relation between the areas of circle and square.

11. The perimeter of a square circumscribing a circle of radius 'a' unit is

(a) 2 units

(b) 4a units

(c) 8a units

(d) 16a Units

12. If the sum of circumferences of two circles with radii R₁ and R₂ is equal to circumference of a circle of radius R₁ then

(a) R₁ + R₂ = R

(b) R₁ + R₂ > R

(d) can't say

13. The area of the largest triangle that can be inscribed in a semicircle of radius r is

(a) r²

b) 2r²

(d) 2r³

14. The area (in cm²) of the circle that can be inscribed in a square of side 8 cm is

(a) 64π

(b) 16π

(d) 32π

15. If the perimeter of a circle is numerically equal to twice its circumference, then the diameter of the circle is

(a) 22:7

(b) 14 : 11

(d) 11: 14

16. If the area of the circle is numerically equal to twice its circumference, then the diameter of the circle is

(a) 4 units

(b) n units

(c) 8 units

(d) 2 units

17. The area of the sector of a circle with radius 6cm and of angle 60⁰ is

(a) 9.42 cm²

(b) 37.68 cm²

(C) 18.84 cm²

(d) 19.84 cm²

18. The length of the minute hard of o clock is 14 cm. The area swept by the minute hand in 5 minutes is

(a) 153.9 cm²

(b)102.6 cm²

(c)51.3 cm²

(d) 205.2 cm²

19. The diameter of a circle whose area is equal to the sun of the areas of the two circles of radii 24cm and 7cm is

(a) 31 cm

(b) 25 cm

(d) 50 cm

20. The area of the largest square that can be inscribed in a circle of radius 12cm is

(a) 24 cm²

(b) 249 cm²

(d) 19√2 cm²