Construction of Tangents to a Circle: A circle is a collection of all points in a plane that is at a constant distance from a fixed point. The fixed point is called the centre, and the constant distance is known as 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. Let us study how to construct a tangent to a given circle and construct tangents to a circle from some given points.

Construction of Tangents to a Circle

Here we have provided the detailed step-by-step on the construction of various tangents of a circle.

Secant

Consider a circle \(c\) and a line \(l\) in a plane when the line \(l\) intersects the circle in two distinct points since a circle cannot pass through three non-collinear points. So, the line intersects the circle in two points only. In such a case, the line \(l\) is called a secant of the circle. So, a line that intersects a circle in two distinct points is called a secant of a circle.

Tangent

Consider the below figure in which the line \(PQ\) is passing through a single point \(A\) on the circle.

The only point \(A\) is common to the line \(PQ\) and the circle. Here, the line \(PQ\) is called the tangent to the circle. So, a tangent to a circle is a line that intersects the circle at exactly one point. This point is called the point of contact of the tangent, and the line is said to touch the circle at this point.

The word tangent is derived from the Latin word Tangere which means to touch. The point of contact is the only point common to the tangent and the circle, and every other point on the tangent lie outside the circle. Thus, of all the points on a tangent to a circle, the point of contact is nearest to the centre of the circle.

The length of the tangent segment from the external point \(P\) to the point of contact with the circle is called the length of the tangent from the point \(P\) to the circle.

Number of Tangents to a Circle

i). No tangents are passing through a point lying inside the circle.
ii). There is only one tangent passing through a point lying on a circle.
iii). There are only two tangents through a point lying outside a circle. In the below figure, \(PA\) and \(PA’\) are two tangents from a point \(P\) lying outside the circle.

Properties of a Tangent

1. The tangent at any point of a circle will be \({90^ \circ }\) to the radius through the point of contact.
2. A line drawn through the end of a radius and the angle between the radius and the line is \({90^ \circ }\) is a tangent to a circle.
3. The measures of tangents drawn from an external point to a circle are equal.
4. If two tangents are drawn from an external point then,
   • – The tangents subtend equal angles at the centre, and
   • – The tangents are equally inclined to the line segment joining the centre to that point
5. In two concentric circles, the chord of the larger circle, which touches the smaller circle, is bisected at the point of contact.
6. The tangents drawn at the finishing point of a diameter of a circle are parallel.
7. The line segment joining the points of contact of two parallel tangents to a circle is the circle’s diameter.
8. The sum of the angle between two tangents drawn from an external point to a circle and the angle subtended by the segments joining the contact points to the centre is \({180^ \circ }.\)
9. There is only one tangent at any point on the circumference of a circle.
10. The perpendicular at the point of contact of the tangent to a circle passes through the centre.

Construction of Tangents to a Circle

If a point lies inside a circle, there cannot be a tangent to the circle through that point. However, if a point lies on the circle, there is only one tangent to the circle, and it is perpendicular to the radius through this point.

Also, if a point lies outside the circle, there will be two tangents to the circle from this point.

Here, we shall learn the construction of tangents to a circle when the centre is known and when its centre is unknown.

Construction of a Tangent to a Circle at a Given Point

Construction of a Tangent to a Circle When its Centre is Known

Steps of construction:

1. Take a point \(O\) and draw a circle of the given radius.
2. Take a point \(P\) on the circle.
3. Join \(OP\)
4. Construct \(\angle OPT = {90^ \circ }.\)
5. Produce \(TP\) to \(T’\) to get \(TPT’\) as the required tangent.

Construction of a Tangent to a Circle at a Given Point When its Centre is Unknown

Steps of construction:

1. Draw any chord \(PQ\) through the given point \(P\) on the circle.
2. Join \(P\) and \(Q\) to a point \(R\) either in the major arc or in the minor arc.
3. Construct \(\angle QPY\) equal to \(\angle PRQ\) and on the opposite side of the chord \(PQ.\)
4. Produce \(YP\) to \(X\) to get \(YPX\) as the required tangent.

Construction of Tangents to a Circle from an External Point

Construction of Tangents to a Circle from an External Point When the Centre is Known

Steps of construction:

1. Join the centre \(O\) of the circle to the given external point \(P,\) i.e., Join \(OP.\)
2. Draw perpendicular bisector of \(OP,\) intersecting \(OP.\) at \(Q.\)
3. Taking \(Q\) as the centre and \(OQ = PQ\) as radius, draw a circle to intersect the given circle at \(T\) and \(T’.\)
4. Join \(PT.\) and \(PT’\) to get the required tangents as \(PT.\) and \(PT’.\)

Construction of Tangents to a Circle from an External Point When its Centre is Unknown

Steps of Construction:

1. Let \(P\) be the external point from where the tangents are to be drawn to the given circle. Through \(P\) draw a secant \(PAB\) to intersect the circle at \(A.\)
2. Produce \(AP\) to a point \(C\) such that \(AP = PC,\) i.e., \(P\) is the mid-point of \(AC.\)
3. Draw a perpendicular bisector of \(BC.\) Draw a semi-circle with \(\frac{{BC}}{2}\) as radius or \(BC\) as diameter.
4. Draw \(PD \bot CB,\) intersecting the semi-circle at \(D.\)
5. With \(P\) as centre and \(PD\) as radius draw arcs to intersect given the circle at \(T\) and \(T’.\)
6. Join \(PT\) and \(T’.\) which are the required tangents.

Solved Examples – Construction of Tangents to a Circle

Q.1 Draw a circle of radius \(3\,{\rm{cm}}.\) Take a point at a distance of \(5.5\,{\rm{cm}}\) from the centre of the circle. From point \(P,\) draw two tangents to the circle.
Ans: Steps of construction:
1. Take a point \(O\) and draw a circle of radius \(3\,{\rm{cm}}.\)
2. Mark a point \(P\) at a distance of \(5.5\,{\rm{cm}}\) from the centre \(O\) and join \(OP.\)
3. Draw the perpendicular bisector of \(OP,\) intersecting \(OP\) at \(Q.\)
4. Taking \(Q\) as centre and \({\rm{OQ = PQ}}\) as radius, draw a circle to intersect the given circle at \(T\) and \(T’.\)
5. Join \(PT\) and \(PT’\) to get the required tangents.

Q.2. Draw a circle of radius \(4\,{\rm{cm}}\) with the centre \(O.\) Draw a diameter \(POQ,\) through \(P\) or \(Q\) draw a tangent to the circle.
Ans: Steps of construction:
1. Taking \(O\) as centre and radius equal to \(4\,{\rm{cm,}}\) draw a circle.
2. Draw diameter of \(POQ.\)
3. Construct \(\angle PQT = {90^ \circ }.\)
4. Produce \(TQ\) to \(T’\) to obtain the required tangent \(TQT’.\)

Q.3. Draw a circle of radius \(4\,{\rm{cm}}.\) Take a point \(P\) on it. Without using the centre of the circle, draw a tangent to the circle at point \(P\)
Ans: Steps of construction:
1. Draw a circle of radius \(4\,{\rm{cm}}.\)
2. Draw any chord \(PQ\) through the given point \(P\) on the circle.
3. Take a point \(R\) on the circle and join \(P\) and \(Q\) to a point \(R.\)
4. Construct \(\angle QPY = \angle PRQ\) and on the opposite side of the chord \(PQ.\)
5. Produce \(YP\) to \(X\) to get \(YPX\) as the required tangent.

Q.4. Draw a circle of radius \(3\,{\rm{cm}}.\) Draw a tangent to this circle, making an angle of \({30^ \circ }\) with a line passing through the centre.
Ans: Steps of construction:
1. Draw a circle of radius \(3\,{\rm{cm}}{\rm{.}}\)
2. Draw a radius \(OA\) of this circle and produce it to \(B.\)
3. Construct an angle \(\angle AOP\) equal to the complement of \({30^ \circ },\) i.e., equal to \({60^ \circ }.\)
4. Draw a perpendicular to \(OP\) at \(P\) which intersects \(OA\) produced at \(Q.PQ\) is the desired tangent such that \(\angle OQP = {30^ \circ }.\)

Q.5. Draw a pair of tangents to a circle of radius \(5\,{\rm{cm,}}\) which are inclined to each at an angle of \({60^ \circ }.\)
Ans: Steps of construction:
1. Take a point \(O\) and draw a circle with centre \(O\) and radius \(OA = 5\,{\rm{cm}}.\)
2. At \(O\) construct radii \(OA\) and \(OB\) such that to \(\angle AOB\) equal \({120^ \circ },\) i.e., the supplement of the angle between the tangents.
3. Draw perpendiculars to \(OA\) and \(OB\) at \(A\) and \(B\) respectively. Suppose these perpendiculars intersect at \(P.\) Then, \(PA\) and \(PB\) are required tangents.

Summary

In the above article, we learnt the definition of secant, tangent, and we learnt about the number of tangents to a circle and the results on a tangent. Also, we have learnt how to construct the tangents to a given circle.

FAQs

Q.1.  How many constructions of tangents to a circle are possible?
Ans: We can draw two tangents to a circle from a point outside of the circle.

Q.2. How do you construct a tangent to a circle through a point outside the circle?
Ans: Steps of construction:
1. Join the centre \(O\) of the circle to the given external point \(P,\) i.e., Join \(OP\)
2. Draw right bisector of \(OP,\) intersecting \(OP\) at \(Q.\)
3. Taking \(Q\) as the centre and \(OQ = PQ\) as radius, draw a circle to intersect the given circle at \(T\) and \(T’.\)
4. Join \(PT\) and \(PT’\) to get the required tangents as \(PT\) and \(PT’.\)

Q.3. Define a tangent of a circle?
Ans: A line meeting a circle only in one point is called a tangent to the circle at that point.

Q.4. Define a secant of a circle?
Ans: A line that intersects a circle in two distinct points is called a secant to the circle.

Q.5. How do you construct a tangent from a point on the circle?
Ans: Steps of construction:
1. Take a point \(O\) and draw a circle of the given radius.
2. Take a point \(P\) on the circle.
3. Join \(OP.\)
4. Construct \(\angle OPT = {90^ \circ }\)
5. Produce \(TP\) to \(T’\) to get \(TPT’\) as the required tangent