Number of Tangents from a Point on a Circle

A circle is a collection of all the points in a plane that are at a constant distance from a particular point. This distance is called the radius of the circle and the fixed point is called the centre.  A straight line and a circle can co-exist in three ways, one can be a straight line with no intersection to the circumference of the circle, there always exists some space between the circle and the line. The second type is when the line lies in such a way that it passes touching the circumference of the circle, known as the Tangents of the circle. The third type is when the line intersects the circumference at two places and is known as a Secant. They are shown below,

1. No Intersection
2. Tangent
3. Secant

A secant intersects the circle at two points while a tangent intersects at only one point. Let's study tangents and their properties in detail. 

Tangents

A line intersecting circle at only one point is called a tangent. A circle has infinitely many possible tangents. The common point of tangent and the circle is called the point of contact. 

Let's see some properties of tangents through theorems. 

Number of Tangents from a Point to a Circle 

It is already known that a Tangent is defined as the line that passes touching the circle's circumference. To get the idea for this, we need to perform some experiments, There are three cases that can be obtained through the different experiments.

Case 1: Tangent from inside a Circle

Let's try to draw a tangent from inside a circle. Let's assume a point P inside the circle and try to draw tangents. We observe that it is not possible to draw tangents from inside the circle. 

Case 2: A point on the circle

We can see that in the figure at the beginning of the article that we can draw a tangent from any point on the circle. In fact, there can be an infinite number of tangents on a circle.

Case 3: A point outside the circle

We can see in the figure that from a point outside the circle, we can draw two tangents to it. 

So, to summarize both the cases:  

1. There is no tangent to a circle from a point inside the circle.
2. There is one tangent to a circle from a point that is on the circle.
3. There are two tangents possible to a circle from a point that is outside the circle.

Properties of Tangents 

Theorem 1:

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

Proof: 

Let's assume a circle with centre O and a tangent XY to circle. 

Let's assume any point Q on the line XY and join the point of contact with centre. The diagram will look like this, 

Now the point Q when joined to centre forms OQ, if it is extended it will become a secant not tangent. Now we can see that, 

OQ > OP 

This is true when Q is any point on the line XY except the point of contact (P). P is the point on line XY whose distance is shortest from the centre O. Thus, OP must be perpendicular to XY. 

Hence, Proved. 

This theorem allows us to conclude some other properties also: 

1. At any point in the circle, there can be only one tangent.
2. The line joining the point of contact and centre is perpendicular to the tangent. Thus, it is also called normal to the tangent.

Theorem 2: 

A line drawn through the end of the radius and perpendicular to it is a tangent to the circle. 

Proof: 

Let's assume a circle with centre O in which OP is the radius. A line AB goes through P such OP is perpendicular to AB. 

Now, take a point Q online AB. We know that the distance of Q is shortest from O when Q = P. In every other case,

OQ > OP and Q lies outside the circle. That means AB meets the circle at only one point P. Thus AB is tangent to the circle. 

Question 1: In the diagram given below. Two tangents are drawn from an external point which is 41 cm from the centre. The radius of the circle is given as 9 cm. Find the area of the quadrilateral AOBC. 

Solution: 

We know from the previous theorem that line joining the centre from point of contact is perpendicular to the tangent. This makes OAC and OBC right-angled triangles. 

So now,

Area of quadrilateral AOBC = Area of triangle AOC + Area of triangle BOC. 

Area of triangle AOC = [Tex]\frac{1}{2} \times base \times height = \frac{1}{2} \times 40 \times 9 = 180  [/Tex] sq units. 

Since both the triangles are congruent, both have the same area. Thus, area of BOC = 180 sq units.

Area of quadrilateral AOBC = 180 + 180 

                                            = 360 sq units. 

Question 2: Let's assume a tangent AB at point A on the circle whose radius is 3cm. The distance of point B from the centre O is 5cm. Find out the length of AB. 

Solution: 

This is also an application of the theorem studied above, let's make a diagram first. 

We can see a right-angled triangle AOC here ,

OC² = OA² + AC² 

52 = 32 + AC² 

25 = 9 + AC²

16 = AC²

AC = 4cm 

Theorem 3: 

The lengths of tangents drawn from an external point to a circle are equal.

Proof: 

Let's assume a circle with centre O, a point C lying outside the circle and the tangents from that point to the circle. AC and BC are the tangents from the point. Our goal is to prove AC = BC.

Let's join OA and OB and consider the two triangles OAC and OBC. 

1. OC is common.
2. ∠OAC = ∠OBC (Right angled triangle)
3. OA = OB (Radii of the circle)

Using RHS property we can say that these two triangles are congruent. Thus, AC = BC. 

Theorem 4: 

If two tangents are drawn from an external point then

1. They subtend an equal angle at the centre, and 
2. They are equally inclined to the line segment joining the centre to that point. 

Proof:

In the given figure, we need to prove that 

∠POA = ∠POB and ∠OPA = ∠OPB.

Let us consider the two triangles, POA and POB. 

PA = PB (By previous theorem) 

OA = OB (radii of the circle) 

OP = OP (Common) 

Thus, these two triangles are congruent. [by SSS] 

Hence, ∠POA = ∠POB and ∠OPA = ∠OPB.

Sample Problems

Question 1: In the given figure, AC and BC are the two tangents drawn from point C. Prove that 2∠OAB = ∠ACB.

Solution: 

We know from the previous theorem that, AC = BC. This concludes that triangle ABC is an isosceles triangle. 

We also know that ∠OAC = 90°. So, 

∠BAC = 90° - ∠OAB

In triangle BAC

∠BAC + ∠ABC + ∠ACB = 180°

2∠BAC + ∠ACB = 180°

2(90° - ∠OAB) + ∠ACB = 180°

180° - 2∠OAB + ∠ACB = 180°

∠ACB = 2∠OAB 

Question 2: There is a circle inscribed in a quadrilateral PQRS, prove that PQ + RS = PS + QR. 

Solution: 

A circle is inscribed inside the quadrilateral PQRS. Notice that the sides of the quadrilateral are actually tangents to the circle. 

PA = PB, 

BQ = QC 

DR = RC 

SA = SD 

We need to prove PQ + RS = PS + QR.

Taking the L.H.S, 

PQ + RS 

⇒ PB + BQ + DR + DS 

⇒ PA + CQ + RC + AS (From the relations stated above)

⇒ (PA + AS) + (CQ + RC) 

⇒ PS + QR 

Hence Proved

Question 3: In concentric circles, prove that chord of the larger circle which touches the smaller circle is bisected at the point of contact. 

Answer: 

Let's say C₁ and C₂ are two concentric circles. Centre is O and AB is the chord of larger circle. From the previous theorems we know that, OP is perpendicular to AB. As we know from the properties of circle, that perpendicular from the centre bisects the chord. 

AB is chord to larger circle C₁ and OP is perpendicular to it. Thus is bisects the chord. 

Properties of Tangents

A tangent is defined as a straight line that intersects a circle exactly once. The following are some of the key properties of tangents:

1. A tangent to a circle is perpendicular to the radius at the point of contact.
2. The lengths of tangents drawn from an external point to a circle are equal.
3. When two tangents are drawn from an external point, they subtend equal angles in the circle's center

Theorems on Tangents

Theorem: A Line Drawn Through the End of the Radius and Perpendicular to It is a Tangent to the Circle

Statement: If a line is drawn through the endpoint of a radius and perpendicular to it, the line is a tangent to the circle.

Solution:

Consider a circle with center O and let P be a point on the circle such that OP is the radius.

Draw a line AB through P such that OP is perpendicular to AB.

Assume there is another point Q on line AB (other than P).

Since Q lies on the line AB and is not on the circle, the distance OQ is greater than OP (which is the radius).

OQ > OP

This implies that the line AB touches the circle at only one point P, and hence AB is a tangent to the circle.

Theorem: The Lengths of Tangents Drawn from an External Point to a Circle are Equal

Statement: If two tangents are drawn from an external point to a circle, then the lengths of the tangents are equal.

Solution:

Consider a circle with center O and a point C outside the circle. Let AC and BC be the two tangents drawn from point C to the circle, touching it at points A and B respectively.

Join OA, OB, and OC.

In triangles OAC and OBC:

OA = OB (Radii of the circle)

OC is common

Angle OAC = Angle OBC (Each is 90 degrees as the radius is perpendicular to the tangent)

By the RHS (Right angle-Hypotenuse-Side) criterion of congruence, triangles OAC and OBC are congruent.

This implies:

AC = BC

Theorem: If Two Tangents are Drawn from an External Point, They Subtend Equal Angles at the Center and are Equally Inclined to the Line Joining the Center to That Point

Statement: If two tangents are drawn from an external point to a circle, they subtend equal angles at the center, and they are equally inclined to the line segment joining the center to that point

Solution:

Consider a circle with center O and an external point P. Let PA and PB be the tangents drawn from P to the circle, touching the circle at points A and B respectively.

Join OA, OB, and OP.

In triangles OPA and OPB:

PA = PB (From the previous theorem)

OA = OB (Radii of the circle)

OP is common

By the SSS (Side-Side-Side) criterion of congruence, triangles OPA and OPB are congruent.

This implies:

∠POA = ∠POB

∠OPA = ∠OPB

Practice Problems

1. A tangent AB is drawn to a circle with center O and radius 5 cm from a point A. If OA = 13 cm, then find the length of the tangent AB.

2. Two tangents PA and PB are drawn from a point P outside a circle. If the distance from P to O is 10 cm and the radius of the circle is 6 cm, find the length of each tangent.

3. In a quadrilateral ABCD, a circle is inscribed, and the sides AB, BC, CD, and DA are tangents to this circle. Prove that AB + CD = BC + DA.

4. Prove that the tangent at any point of a circle is perpendicular to the radius through the point of contact.

5. A chord of length 6 cm is drawn in a circle of radius 4 cm. Determine the distance of the chord from the centre of the circle.

Conclusion

This article explains the concept of tangents to circles and their theorems, which are crucial in geometry for solving circle-related problems. Tangents, such as their perpendicularity to the radius and equality from an external point, aid in solving various geometric problems, enhancing understanding through practice problems.