Tangent and Secant Properties of a Circle: Circle is the collection of all points in a plane at a constant distance from a fixed point. The fixed point is known as the centre, and the constant distance is the circle’s radius. A line that intersects a circle in two distinct points is called a secant to the circle.

A line meeting a circle only in one point is called a tangent to the circle at that point. In this article, we will talk about the relative position of the circle and a line in the same plane. We will also examine the existence of tangents to a circle and their properties.

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Circle and Its Parts

A circle is the locus of a point that moves in a plane so that its distance from a fixed point in the same plane always remains constant.

As you can see in the diagram, \(O\) is that fixed point called the centre and \(OA,\) or \(OB\) is the fixed distance called the circle’s radius. \(CD\) is a chord of the circle, and \(AB\) is a diameter, which is the longest chord. The region bounded by the two radii \(OP\) and \(OQ\) and the arc \(PQ\) is called a sector, and the region enclosed by chord \(PQ\) and arc \(PQ\) is called a segment.

Secant of a Circle

A line intersecting a circle in two distinct points is the secant of the circle. In the given diagram \(AB\) is a secant of the circle.

Tangent: A line meeting a circle in only one point is called the tangent to the circle. In the given diagram, line \(AB\) is tangent to the circle at \(P.\)

Learn About Properties of Circle

Point of contact: The point \(P,\) where a tangent to the circle meets the circle, is called the point of contact of the tangent. In the given diagram, \(AB\) touches the circle at \(P.\) Therefore, \(P\) is the point of contact of the tangent.

Intersecting Secant-Tangent Theorem

In the moving bicycle, the wheels move along a line that is tangent to the circular wheel. The spokes of the wheel are along its radii, and each spoke through the point of contact of the wheel, and the ground is perpendicular to the line along which the wheel moves. This property of tangent to a circle is established in the following theorem:

Theorem 1: The tangent at any point of a circle is perpendicular to the radius through the point of contact.

Given: A circle with centre \(O\) and a tangent \(AB\) at a point \(P\) of the circle.

To prove: \(OP⊥AB\)

Construction: We take a point \(Q,\) other than \(P,\) on \(AB\) and join \(OQ.\)

Proof: Since \(Q\) is a point on the tangent \(AB,\) other than the point of contact \(P,\)

\(∴ Q\) lies outside the circle

Let \(OQ\) intersect the circle at \(R.\)

Now,  \(OQ=OR+RQ\)
\(⇒OQ>OR\)
\(⇒OR<OQ\)
\(⇒OP<OQ\) [\(∴OP=OR=\)radius of the circle]

Thus, \(OP\) is shorter than any other line segment joining \(O\) to any point of \(AB,\) other than \(P.\)

i.e., \(OP\) is the shortest distance between the point \(O\) and the line \(AB.\)

But, the shortest distance between a point and a line is the perpendicular distance from the point to the line.

\(∴ OP⊥AB\)

Theorem 2: (Converse of theorem 1): A line drawn through the end of a radius and perpendicular to it is a tangent to the circle.

Given: A circle having centre \(O\) of which \(OP\) is a radius and \(AB\) is a line through \(P\) such that \(OP⊥AB.\)

To prove: \(AB\) is tangent to the circle at the point \(P.\)

Construction: We take a point \(Q,\) different from \(P,\) on \(AB\) and join \(OQ.\)

Proof: We know that the perpendicular distance between points to lines is the shortest distance between them.

\(∴ OP ⊥AB\) [\(∴OP\) is the shortest distance from \(O\) to \(AB\)]
\(⇒OP<OQ\)
\(⇒OQ>OP\) 
\(⇒Q\) lies outside the circle [\(∴OP\) is the radius and \(OP<OQ\)]

Thus, every point on \(AB,\) other than \(P,\) lies outside the circle.

\(∴ AB\) meets the circle at only one point \(P.\)

Hence, AB is the tangent to the circle at the point \(P.\)

Note: At any point on a circle, there can be one and only one tangent.

The line perpendicular to a tangent at the point of contact is known as the normal to the circle at the point of control.

Tangent and Secant Formula

Suppose a tangent segment and the secant segment are drawn to a circle from an exterior point. Here, the square of the measure of the tangent segment is equal to the product of the measures of the secant segment and its external secant segment. They intersect at point \(U.\)

So, \(U{V^2} = UX \cdot UY\)

If a secant and a tangent of a circle are drawn from a point outside the circle, then;

\({\rm{Lengths}}\,{\rm{of}}\,{\rm{the}}\,{\rm{secant \times external}}\,{\rm{segment = (length}}\,{\rm{of}}\,{\rm{the}}\,{\rm{tangent}}\,{\rm{segment}}{{\rm{)}}^{\rm{2}}}\)

What are the Tangent and Secant Properties of a Circle?

The tangent properties are given below: 

1. One tangent can touch the circle at only a single point of the circle.
2. A tangent never crosses the circle, that means it cannot pass through the circle.
3. A tangent never intersects the circle a couple of points.
4. The tangent line is perpendicular to the radius of the circle.
5. The point where the intersection occurs is called the point of tangency.

The secant properties are given below: 

1. A secant is a line that intersects a circle in exactly two points.

Learn Construction of Tangents to a Circle

Solved Examples: Tangent and Secant Properties of a Circle

Q.1. From a point \(P,\) \({\rm{10}}\,{\rm{cm}}\) away from the centre of a circle, a tangent \(PT\) of length \({\rm{8}}\,{\rm{cm}}\) is drawn. Find the radius of the circle.
Ans: Let \(O\) be the centre of the given circle and let \(P\) be a point such that \(OP{\rm{ = 10}}\,{\rm{cm}}{\rm{.}}\)
Let \(PT\) be the tangent to the circle such that \(PT{\rm{ = 8}}\,{\rm{cm}}{\rm{.}}\)
We join \(OT.\)
Since \(PT\) is a tangent to the circle at \(T\) and \(OT\) is the radius through \(T.\)

\(∴ OT ⊥PT\)
In \(\Delta OTP,\angle OTP = {90^{\rm{o}}}\)
\(\therefore O{P^2} = O{T^2} + P{T^2}\) [By Pythagoras theorem]
\( \Rightarrow OT = \sqrt {O{P^2} – P{T^2}} = \sqrt {{{10}^2} – {8^2}} \;{\rm{cm}} = \sqrt {36} \;{\rm{cm}} = 6\;{\rm{cm}}\)
Hence, the radius of the circle is \({\rm{6}}\,{\rm{cm}}{\rm{.}}\)

Q.2. A tangent \(PQ\) at a point \(P\) of a circle of radius \({\rm{5}}\,{\rm{cm}}\) meets a line through the centre O at a point \(Q\) so that \(OQ{\rm{ = 12}}\,{\rm{cm}}{\rm{.}}\). Find the length of \(PQ\).

Ans: Since tangent at any point to a circle is perpendicular to the radius through the point of contact,
\(∴ OP⊥PQ\)
In the right \(\angle OPQ,\angle OPQ = {90^{\rm{o}}}\)
\(\therefore O{Q^2} = O{P^2} + P{Q^2}\) [By Pythagoras theorem]
\( \Rightarrow {12^2} = {5^2} + P{Q^2}\)
\( \Rightarrow P{Q^2} = 144 – 25 = 119\)
\( \Rightarrow PQ = \sqrt {119} \;{\rm{cm}}\)

Q.3. The length of a tangent from a point \(A\) at a distance \(5\;{\rm{cm}}\) from the centre of the circle is \(4\;{\rm{cm}}.\) Find the radius of the circle.
Ans: Given: A circle with centre \(O\)
\(OA = 5\;{\rm{cm}},AP = 4\;{\rm{cm}}\)
To find: \(OP\)
In right \(\Delta APO,\angle OPA = {90^{\rm{o}}}\)
\(\therefore O{A^2} = A{P^2} + O{P^2}\) [By Pythagoras theorem]
\( \Rightarrow {5^2} = {4^2} + O{P^2}\)
\( \Rightarrow O{P^2} = 25 – 16\)
\( \Rightarrow O{P^2} = 9\)
\( \Rightarrow OP = 3\;{\rm{cm}}\)

Hence, the radius of the circle is \(3\;{\rm{cm}}.\)

Q.4. If tangents \(PA\) and \(PB\) from a point \(P\) to a circle with a centre \(O\) are inclined to each other at an angle of \({80^{\rm{o}}},\) then find \(∠POA.\)
Ans: Given: \(PA\) and \(PB\) are tangents to a circle with \(O.\)
Also \(\angle OAP = {80^{\rm{o}}}\)
To find: \(∠POA\)

Since \(OP\) bisects \(∠APB\)
\(\therefore \angle APO = \frac{1}{2}\angle APB\)
\( \Rightarrow \angle APO = \frac{1}{2}\left( {{{80}^{\rm{o}}}} \right) = {40^{\rm{o}}}\)
We have, \(∠OA⊥AP\) [radius is perpendicular to tangents]
\( \Rightarrow \angle OAP = {90^{\rm{o}}}\)
Now, in \(△PAO,\)
\(\angle OAP + \angle APO + \angle POA = {180^{\rm{o}}}\)
Sum of angles of a triangle\( = {180^{\rm{o}}}\)
\( \Rightarrow {90^{\rm{o}}} + {40^{\rm{o}}} + \angle POA = {180^{\rm{o}}}\)
\( \Rightarrow {130^{\rm{o}}} + \angle POA = {180^{\rm{o}}}\)
\( \Rightarrow \angle POA = {180^{\rm{o}}} – {130^{\rm{o}}}\)
\( \Rightarrow \angle POA = {50^{\rm{o}}}\)

Q.5. If \(TP\) and \(TQ\) are the two tangents to a circle with centre \(O\) so that \(\angle POQ = {110^{\rm{o}}},\) find \(\angle PTQ.\)

Ans: Given: \(TP\) and \(TQ\) are the two tangents to a circle with centre \(O\) and \(\angle POQ = {110^{\rm{o}}}.\)
To find: \(∠PTQ\)
Since \(TP\) and \(TQ\) are tangents to the circle at \(P\) and \(Q\) respectively.
\(\therefore \angle OPT = \angle OQT = {90^{\rm{o}}}\)
\( \Rightarrow \angle POQ + \angle OQT + \angle QTP + \angle TPO = {360^{\rm{o}}}\)
[\(Q\) Sum of angles of a quadrilateral is \({360^{\rm{o}}}\)]
\( \Rightarrow {110^{\rm{o}}} + {90^{\rm{o}}} + \angle QTP + {90^{\rm{o}}} = {360^{\rm{o}}}\)
\( \Rightarrow {290^{\rm{o}}} + \angle QTP = {360^{\rm{o}}}\)
\( \Rightarrow \angle QTP = {360^{\rm{o}}} – {290^{\rm{o}}} = {70^{\rm{o}}}\)

Summary

In the given article, we have discussed the tangent and secant properties of a circle followed by the secant of a circle. Then we talked about intersecting secant-tangent theorem, which includes
Theorem 1: The tangent at any point of a circle is perpendicular to the radius through the point of contact, and
Theorem 2: (Converse of theorem 1): A line drawn through the end of a radius and perpendicular to it is a tangent to the circle.

We glanced at the properties of tangent and secant of a circle. We have provided solved examples along with a few FAQs.

Frequently Asked Questions (FAQs)

Q.1. What are the properties of a tangent to the circle?
Ans: One tangent can touch a circle at only one point of the circle. A tangent never crosses a circle, which means it cannot pass through the circle. A tangent never intersects the circle at the two points. The tangent line is perpendicular to the radius of a circle.

Q.2. What is tangent secant property?
Ans: If the tangent segment and a secant segment are drawn to the circle from an exterior point, then the square of the measure of a tangent segment is similar to the product of the measures of the secant segment and its external secant segment.

Q.3. What are the tangent properties?
Ans: The tangent properties are given below:
One tangent can touch the circle at only a single point of the circle.
A tangent never crosses the circle, that means it cannot pass through the circle.
A tangent never intersects the circle a couple of points.
The tangent line is perpendicular to the radius of the circle.

Q.4. How do you find the tangent and secant of a circle?
Ans: If a line is touching at only one point in a circle, it’s is a tangent to the circle. If the line touches or intersects the circle at two different points, then it is the secant to the circle.

Q.5. What do you mean by the tangent to a circle?
Ans: A tangent to a circle is a straight line that touches the circle at only one point.

We hope this detailed article on the tangent and secant properties of a circle helped you in your studies. If you have any doubts, queries or suggestions regarding this article, feel to ask us in the comment section and we will be more than happy to assist you. Happy learning!