Circles: Properties, Formulas, and Practical Applications

Circles

A circle is a collection of all those points in a plane that are at a given constant distance from a given fixed point in the plane. The fixed point is called the centre and the given constant distance is called the radius of the circle.

1.0Intersection of a circle and a line

Consider a circle with centre 0 and radius $r$ and a line $PQ$ in a plane. We find that there are three different positions, a line can take with respect to the circle as shown in the following figures. (1) The line PQ does not intersect the circle. In figure the line $PQ$ and the circle have no common point. In this case $PQ$ is called a nonintersecting line with respect to the circle.

(2) The line $PQ$ intersect the circle in more than one point. In fig, there are two common points $A$ and $B$ between the line $PQ$ and the circle and we call the line $PQ$ as a secant of the circle.

Circles having the same centre but with different radii are said to be concentric circles.
There is a difference between a chord and a secant. Chord is a line segment joining any two points on a circle.

(3) The line intersects the circle in a single point i.e. the line intersects the circle in only one point. In fig we can verify that there is only one point ' A ' which is common to the line PQ in the given circle. In this case the line is called a tangent to the circle.

Secant A secant is a straight line that cuts the circumference of the circle at two distinct (different) points i.e., if a circle and a line have two common points then the line is said to be secant to the circle.

The measurement of the boundary of the circle is called circumference of the circle.

Tangent A tangent is a straight line that meets the circle at one and only one point. This point ' A ' is called point of contact or point of tangency as shown in figure.

Tangent as a limiting case of a secant

In the figure, the secant $ℓ$ cuts the circle at A and B . If this secant $ℓ$ is turned around the point A , keeping $A$ fixed then $B$ moves on the circumference closer to $A$ . In the limiting position, $B$ coincides with A. The secant $ℓ$ becomes the tangent at A. Tangent to a circle is a secant when the two end points of its corresponding chord coincide.

In the figure, $ℓ$ is a secant which cuts the circle at A and B . If the secant is moved parallel to itself away from the centre, then the points $A$ and $B$ come closer and closer to each other. In the limiting position, they coincide into a single point at A, the secant $ℓ$ becomes the tangent at A. Thus, a tangent line is the limiting case of a secant when the two points of intersection of the secant and a circle coincide with the point A . The point A is called the point of contact of the tangent. The line $ℓ$ touches the circle at the point A. i.e., the common point of the tangent and the circle is called the point of contact.

Point of tangency/ Point of contact

The line containing the radius through the point of contact is called normal to the circle at the point.

2.0Number of tangents to a circle from a point

If a point A lies inside a circle, no line passing through ' A ' can be a tangent to the circle. i.e., No tangent can be drawn from the point $A$ .

If A lies on the circle, then one and only one tangent can be drawn to pass through 'A'. i.e. Exactly one tangent can be drawn through A.

If A lies outside the circle then exactly two tangents can be drawn through ' A '. In the figure, a secant ABC is drawn from a point ' A ' outside the circle, if the secant is turned around A in the clockwise direction, in the limiting position, it becomes a tangent at T. Similarly, if the secant is turned in the anti-clockwise direction, in the limiting position, it becomes a tangent at S . Thus, from a point A outside a circle only two tangents can be drawn. i.e. AS and AT.

3.0Properties of tangent to a circle

Theorem-1 : The tangent at any point of a circle and the radius through the point are perpendicular to each other. Given : A circle with centre $O . AB$ is a tangent to the circle at a point P and OP is the radius through P .

To prove :

r

. Construct : Take a point

Q

, other than

P

, on tangent

A B

. Join

OQ

. Proof:

Q

is a point on tangent

A B

, other than the point

P, S o, Q

will lie outside the circle

∴ 0 Q

will intersect the circle at some point

R

O Q & =O R+Q R \\ O Q & =O P+Q R (\because O P=O R) \\ \Rightarrow \quad O P <O Q (O R=O P=\text { radius. })

Thus,

OP

is shorter than any other line segment joining 0 to any point on

A B

all line segments drawn from 0 to line

A B

, the perpendicular is the shortest distance.

∴ OP ⊥ AB

A pair of tangents drawn at two points of a circle are either parallel or they intersect each other at a point outside the circle.
If two tangents drawn to a circle are parallel to each other, then the line-segment joining their points of contact is a diameter of the circle.
The distance between two parallel tangents to a circle is equal to the diameter of the circle, i.e., twice the radius.
A pair of tangents drawn to a circle at the end points of a diameter of a circle are parallel to each other.
A pair of tangents drawn to a circle at the end points of a chord of the circle, other than a diameter, intersect each other at a point outside the circle.

Corollary 1 : A line drawn through the end point of a radius and perpendicular to it is a tangent to the circle. Given : $O$ is the centre and $r$ be the radius of the circle. $OP$ is a radius of the circle. Line $ℓ$ is drawn through P so that $OP ⊥ ℓ$ .

To prove : Line

ℓ

is tangent to the circle at

P

. Construction : Suppose that the line

ℓ

is not the tangent to the circle at P. Let us draw another straight line

m

which is tangent to the circle at

P

. Take two points

A

and

B

(other than

P

) on the line

ℓ

and two points C and D on m . Proof: As OP

O . AB

(Given)

∠ OPB = 9 0^{\circ}

As

OP ⊥ m

(By theorem)

\Rightarrow ∠ OP D = 9 0^{\circ}

\Rightarrow ∠ OPD = ∠ OPB

(Each

= 9 0^{\circ}

) But a part cannot be equal to whole. This gives contradiction. Hence, our supposition is wrong. Therefore, the line

ℓ

is tangent to the circle at P .

Corollary 2: If $O$ be the centre of a circle and tangents drawn to the circle at the points A and B of the circle intersect each other at P , then $∠ AOB + ∠ APB = 18 0^{\circ}$ .

Proof: As

OA ⊥ PA & OB ⊥ PB

\Rightarrow ∠ OAP = ∠ OBP = 9 0^{\circ}

∠ AOB + ∠ OBP + ∠ OAP + ∠ APB = 36 0^{\circ}

(By theorem 1) (Angle sum property of quadrilateral)

\Rightarrow ∠ AOB + 9 0^{\circ} + 9 0^{\circ} + ∠ APB = 36 0^{\circ}

\Rightarrow ∠ AOB + ∠ APB + 18 0^{\circ} = 36 0^{\circ}

\Rightarrow ∠ AOB + ∠ APB = 36 0^{\circ} - 18 0^{\circ}

\Rightarrow ∠ AOB + ∠ APB = 18 0^{\circ}

A quadrilateral in which the sum of the opposite angle is $18 0^{\circ}$ , then it is a cyclic quadrilateral.

Theorem-2 : If two tangents are drawn to a circle from an exterior point, then (i) the tangents are equal in length (ii) the tangents subtend equal angles at the centre (iii) the tangents are equally inclined to the line joining the exterior point and the centre of the circle. Given : PA and PB are two tangents drawn to a circle with centre 0 , from an exterior point P .

To prove : (i)

PA = PB

(ii)

∠ AOP = ∠ BOP

(iii)

∠ APO = ∠ BPO

. Proof: In

△ A OP

and

△ BOP

\Rightarrow OA = OB

(Radii of the same circle.)

\Rightarrow ∠ OAP = ∠ OBP = 9 0^{\circ}

(Radius through point of contact is perpendicular to the tangent)

\Rightarrow OP = OP

(Common)

∴ △ AOP ≅ △ BOP

(By RHS congruence) Hence, we have (i)

PA = PB

(c.p.c.t.) (ii)

∠ AOP = ∠ BOP

(c.p.c.t.) (iii)

∠ APO = ∠ BPO

. (c.p.c.t.)

Theorem 2 can also be proved by using the Pythagorean theorem as follows:

$PB^{2} = OP^{2} - PB^{2}, OP^{2} - OA^{2} = PA^{2} (∵ ∠ PBO = ∠ PAO = 9 0^{\circ}) \Rightarrow PA^{2} = PB^{2} (∵ OA = OB) \Rightarrow PA = PB$

Corollary 3 : If PA and PB are two tangents from a point to a circle with centre 0 touching it at A and $B$ , prove that $OP$ is perpendicular bisector of $A B$ .

Proof: In $△ ACP$ and $△ BCP$ (i) $PA = PB$ (Lengths of two tangents from P are equal) (ii) $PC = PC$ (Common) (iii) $∠ APC = ∠ BPC$ (By theorem 2) $△ ACP ≅ △ BCP$ (SAS congruency) $A C = BC$ $∠ ACP = ∠ BCP$ (c.p.c.t.) $∴ ∠ ACP = ∠ BCP = 9 0^{\circ}$ (By linear pair) Hence $OP$ is perpendicular bisector of $A B$ .

4.0Common tangents of two circles

Two circles in a plane, either intersect each other at two points or touch each other at a point or they neither intersect nor touch each other.

Common tangents of two non-intersecting and non-touching circles

Here, we observe that in the above fig, there is no common tangent.

Common tangents of two circles which touch each other internally at a point Two circles touch each other internally at C. Here, we have only one common tangent of the two circles as shown in the below figure.

Common tangent of two intersecting circles Two circles intersect each other at two points A and B. Here, PP' and QQ' are the only two common tangents. The case where the two circles are of unequal radii, we find the common tangents PP' and QQ' are not parallel.

Common tangents of two circles which touch each other externally at a point.

Two circles touch each other externally at C . Here,

PP^{'}, QQ^{'}

and AB are the three common tangents drawn to the circles.

5.0Common tangents of two non-intersecting and non-touching circles

Here, we observe that in fig, there are four common tangents $PP^{'}, QQ^{'}, AA^{'}$ and $BB^{'}$ .

Tangents PP' and QQ' are direct common tangents while tangents $AA^{'}$ and $BB^{'}$ are transverse common tangents.

In the given figure 0 is the centre of a circle. $PT$ and $PQ$ are tangents to the circle from an external point P. If $∠ TPQ = 7 0^{\circ}$ , find $∠ TRQ$ .

Numerical Ability 1 A point $A$ is 26 cm away from the centre of a circle and the length of tangent drawn from $A$ to the circle is $24 cm$ . Find the radius of the circle. Solution: Let $O$ be the centre of the circle and let $A$ be a point outside the circle such that $O A = 26 cm$ . Let AT be the tangent to the circle. Then, AT $= 24 cm$ . Join OT. Since the radius through the point of contact is perpendicular to the tangent, we have $∠ OTA = 9 0^{\circ}$ . In right $△ OTA$ , we have:

OT^{2} = OA^{2} - AT^{2}

= [(26)^{2} - (24)^{2}] = (26 + 24) (26 - 24) = 100

.

\Rightarrow OT = 100 = 10 cm

. Hence, the radius of the circle is 10 cm .

Pythagoras theorem: In right angled triangle, square of hypotenuse is equal to the sum of square of other 2 sides.

Numerical Ability 2 In the given figure, $A$ is right-angled at $B$ , in which $A B = 15 cm$ and $BC = 8 cm$ . A circle with centre 0 has been inscribed in $△ A BC$ . Calculate the value of $x$ , the radius of the inscribed circle.

Solution: Let the inscribed circle touch the sides

AB, BC

and CA at

P, Q

and R respectively. Applying Pythagoras theorem on right

△ OTA

, we have

AC^{2} = AB^{2} + BC^{2} = (15)^{2} + (8)^{2} = (225 + 64) = 289

\Rightarrow AC = 289 = 17 cm

. Clearly,

OPBQ

is a square.

[∵ ∠ OPB = 9 0^{\circ}, ∠ PBQ = 9 0^{\circ}, ∠ OQB = 9 0^{\circ}

and

△ A BC

∴ BP = BQ = xcm

. Since the tangents to a circle from an exterior point are equal in length, we have

A R = A P

and

CR = CQ

. Now,

AR = AP = (AB - BP) = (15 - x) cm

CR = CQ = (BC - BQ) = (8 - x) cm

.

∴ AC = AR + CR

\Rightarrow 17 = (15 - x) + (8 - x)

\Rightarrow 2 x = 6

\Rightarrow x = 3

Hence, the radius of the inscribed circle is 3 cm.

Numerical Ability 3 In the given figure, the incircle of $△ A BC$ touches the sides $A B, BC$ and $C A$ at the points $P, Q$ , $R$ respectively.

Show that

A P + BQ + CR = BP + CQ + A R = \frac{1}{2} (

Perimeter of

△ A BC)

Explanation: Since the lengths of two tangents drawn from an external point to a circle are equal, we have

A P = A R, BQ = BP and CR = CQ ∴ A P + BQ + CR = A R + BP + CQ \dots (i) P er im e t er of △ A BC = A B + BC + C A = A P + BP + BQ + CQ + A R + CR = (A P + BQ + CR) + (BP + CQ + A R) = 2 (A P + BQ + CR) ∴ A P + BQ + CR = BP + CQ + A R = \frac{1}{2} (Perimeter of △ A BC) .

Numerical Ability 4 Two circles of radii 25 cm and 9 cm touch each other externally. Find the length of the direct common tangent Solution: Let the two circles with centres A and B and radii 25 cm and 9 cm respectively touch each other externally at a point C .

Then,

AB = AC + CB = (25 + 9) cm = 34 cm

. Let

PQ

be a direct common tangent to the two circles. Join AP and BQ. Then,

AP ⊥ PQ

and

BQ ⊥ PQ

.

[∵

Radius through point of contact is perpendicular to the tangent] Draw, BL

⊥

AP. Then,

P L BQ

is a rectangle. Now,

LP = BQ = 9 cm

and

PQ = BL

.

∴ AL = (AP - LP) = (25 - 9) cm = 16 cm

. From right

△ A L B

, we have

AB^{2} = AL^{2} + BL^{2}

\Rightarrow BL^{2} = AB^{2} - AL^{2} = (34)^{2} - (16)^{2}

= (34 + 16) (34 - 16) = 900

\Rightarrow BL = 900 = 30 cm

. Length of direct common tangent

= d^{2} - (r_{1} - r_{2})^{2}

Where

d =

distance between the centres of the circles

∴ PQ = BL = 30 cm

. Hence, the length of direct common tangent is 30 cm .