## Solid Geometry Chapter 8 Construction Of Tangent To A Circle

**What is tangent to a circle?**

**Definition:** If a straight line intersects any circle on a plane at only one point, i.e. if there exists only one common point of the straight line and the circle, then the straight line is called the tangent to the circle.

For example, in the adjoining figure the straight line AB intersects the circle with the centre at O at only one point P.

So, AB is tangent to the circle with the centre at O.

**Number of tangents**

Only one tangent can be drawn on a point of any circle, i.e:, the number of tangent = 1.

**Common tangent**

If a straight line touches two or more than two circles, then the straight line is called the common tangent of the circles.

**Types of common tangents:** Common tangents can be of three types. Such as,

**WBBSE Solutions for Class 10 Maths**

**Common tangents drawn at the point of intersection:**

If two circles intersect each other internally or externally, then we can draw a common tangent at the point of intersection.

Here the tangent passes through the common point of intersection of the two circles.

**Direct common tangent:**

If a straight line touches both of the disjoint circles, then the straight line is called the common tangent to the circles.

Such as, in the figure above, PQ and RS are both common tangents to the two circles with centres at A and B.

The important characteristic of a direct common tangent is that here the two point of intersection of the two circles lie on the same side of the line segment obtained by joining the centres of the two circles.

The number of direct common tangents is almost 3.

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**Transverse common tangent:**

If a straight line touches transversely both of the disjoint circles, then the straight line is called the transverse common tangent of the two circles.

Such as in the adjoining PQ and RS are both the transverse common tangents to the two circles with centres at A and B.

The important characteristic of transverse common tangent is, here the points of contact of the two circles lie on the opposite sides of the line segment obtained by joining the two centres of the circles.

The number of the transverse common tangents of two disjoint circles is 2.

**Point of contact and Radius through the point of contact:**

The point at which any straight line touches a circle is called the point of contact and the line segment joining the point of contact and the centre of the circle is called the radius through the point of contact.

In the image beside, the tangent PT touches the circle with the centre at O at the point P.

So, here P is D the point of contact and OP is the radius through point of contact P.

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One important characteristic of radius through point of contact is that it is always perpendicular to the tangent at the point of contact.

A theorem related to this you have proved earlier.

∴ OP ⊥ PT, i.e., ∠OPT = 1 right angle or 90°.

The number of transverse common tangents is at most 2.

We shall now discuss different types of construction of tangents to a circle.

**Construction of a tangent to the circle on any point of its circumference.**

Let P be any point on the circumference of the circle with centre at O.

We have to construct a tangent to this circle at P.

**Method of construction:**

Let us join O, P. Let us also draw a perpendicular PT on OP at P. Then PT is the required tangent.

**Proof:** OP is a radius of the circle with the centre at O which intersects the circle at P and OP ⊥ PT.

∴ PT has required tangent to the circle with centre O at point P.

**Construction of a tangent to a circle from any external point of the circle.**

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Let P be an external point of the circle with the centre at O. We have to construct a tangent to the circle from point P.

**Method of construction:**

Let us join O, P. Let us construct the perpendicular bisector CD of OP.

Let CD intersects OP at B. Let us then draw a semi-circle with centre at B and a radius equal to BO or BP.

Let the semicircle intersects the circle with centre at O at the point A. Let us join P, A and PA is extended to T.

Then PT is the required tangent drawn from the external point P to the circle with centre at O.

**Proof:** Let us join O, A.

Since ∠OAP is a semicircular angle of the semicircle with diameter OP,

∴ ∠OAP = 90°, i.e., OA ⊥ PT.

But OA is a radius passing through point of contact A.

∴ PT is a tangent to the circle with centre at O. (Proved)

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**Construction of two tangents to a circle from an external point of the circle.**

Let P be any point of the circle with centre at C. the point P.

**Method of construction:**

- Let us draw a circle with centre at C and with any radius.
- Let us take any external point P of the circle.
- Let us join C and P.
- Let us bisect CP so that D is the mid-point of CP.
- Let us draw another circle with centre at D and with radius equal to CD. Let this circle intersects the circle with centre at C at two points A and B.
- Then, let us join P, A and P, B.

Hence PA and PB are two required tangents to the circle with centre at C.

**Proof:** Let us join C, A.

In the circle with centre at D, ∠PAC is a semicircular angle. ∴ ∠PAC = 90°.

Again, CA is the radius through point of contact. ∴ CA ⊥ PA,

∴ PA is a tangent to the circle with centre at C.

Similarly, it can be proved that PB is another tangent to the circle with centre at C.

Hence PA and PB are the required two tangents to the circle with centre at C from an external point P of the circle. (Proved)

In the following examples various applications of the above constructions are discussed.

## Solid Geometry Chapter 8 Construction Of Tangent To A Circle Examples

**Example 1. Draw a circle of radius 3.2 cm. Then construct a tangent to that circle on any point of that circle.**

Solution:

**Given:**

Radius 3.2 cm

Let r = 3.2 cm and O is the centre of the circle.

Let us draw a circle with centre at O and radius = 3.2 cm.

Let P be a point on the circle. We have to construct a tangent at the point P.

**Method of construction:**

- Let us join O, P.
- Let us draw PT ⊥ OP.
- TP is extended to Q.

Then QPT is the required tangent.

**Example 2. Draw a line segment AB, the length of which is 3 cm. Draw a circle with centre at A and with radius equal to AB. Then construct a tangent to that circle at the point B.**

Solution:

**Given:**

length is 3 cm.

PT is the required tangent to the circle with centre at A.

**Example 3. Construct a circle of radius 2.5 cm. Take a point at a distance of 6.5 cm from the centre of that circle. Then draw a tangent to that circle from that external point and find the length of the tangent by a scale.**

Solution:

**Given:**

Radius 2.5 cm

Here PT is the required tangent and PT = 6 cm.

**Example 4. Construct a circle of radius 2.8 cm. Take a point at a distance of 7.5 cm from the centre of the circle. Draw two tangents to that circle from that external point.**

Solution:

**Given:**

Radius 2.8 cm.

Take a point at a distance of 7.5 cm from the centre of the circle.

Here PQ and PR are the two tangents to the circle with centre at O from the external point P at u distance of 2.8 cm from the centre O.

**Example 5. PQ is a chord of the circle with centre at O. Draw two tangents at P and Q respectively.**

Solution:

Here, PT and QS are the two required tangents at P and Q respectively.

**Example 6. Draw a line segment XY of length 8 cm and taking XY as the diameter, draw a circle. Then construct two tangents to that circle at the points X and Y. Also find the relation between the two tangents.**

Solution: Here, PQ and RS are the two required tangents at the points Y and X respectively.

**Relation between PQ and RS**:

We see that PQ and RS are parallel to each other.

**Example 7. Draw an equilateral triangle of sides 5 cm and then draw a circumcircle of that triangle. Also draw three tangents at A, B and C respectively.**

Solution:

**Given:**

Equilateral triangle of sides 5 cm.

The circle with centre at O is the required circumcircle of the equilateral triangle ABC.

The three tangents at A, B and C are PQ, RS and UV respectively.

**Example 8. Construct an equilateral triangle ABC of sides 5 cm each and then construct its circumcircle. Draw a tangent at A of the circle and then take a point P on it such that AP = 5 cm. Draw another tangent to the circle from the point P and observe minutely at what point of the circle this tangent intersects.**

Solution: We see that another tangent from P touches the circle at the point C.

**Example 9. p is any point on the line segment AB. Draw a perpendicular PQ at O on AB. Draw two circles with centres at A and B and radius equal to AO and BO. Also write what PQ is called with respect to these circles.**

Solution: Here PQ is said to be a direct common tangent with respect to the circles with centres at A and B.

**Example 10. P is any point on the circle with centre at O. Draw a tangent to that circle at P and cut off the part PQ equal to the radius of the circle from that tangent. From the point Q, draw another tangent QR to that circle and find the value of ∠PQR.**

Solution: Here QR is the required another tangent to the circle with centre at O and ∠PQR = 90°.

**Example 11. Construct a circle of radius 2.5 cm. Take any point on the circle and draw a tangent to the circle at that point.**

Solution: Here PQ is the required tangent to the circle with centre at O.

**Example 12. Draw a circle of radius 2 cm. Draw any triangle inside the circle so that the drawn circle be the circumcircle of the triangle. Now, draw three tangents to the circle with centre at O at the three vertices of that triangle.**

Solution:

**Given:**

Radius 2 cm.

Here O is the centre of the circumcircle of ΔABC.

PQ, RS and UV are the three tangents to the circle at the vertices A, B and C respectively.

**Example 13. Draw a circle of radius 3 cm. Take any point at a distance of 5 cm from the centre of that circle and then construct a tangent to the circle from that point.**

Solution:

**Given:**

Radius 3 cm.

Here PT is the required tangent to the circle with centre at O.

**Example 14. Construct the circumcircle by drawing an equilateral triangle of sides 5 cm each. Also draw two tangents to that circle at A and C which intersect each other at Write what type of the quadrilateral ABCP is.**

Solution: Here, O is the centre of the circumcircle of ΔABC.

The tangent EAP at A and the tangent TCP at C intersect each other at P.

∵ AB II PC and AP II BC, ∴ the quadrilateral ABCP is a parallelogram.

**Class 10 Maths Wbbse Solutions**

## Solid Geometry Chapter 8 Construction Of Tangent To A Circle Construction Of Common Tangents

You have already studied about the common tangents of two circles. You have also known that common tangents are of two types—direct common tangents and transverse common tangents.

We shall now discuss how common tangents (direct and transverse) of two circles are drawn.

**Construction of direct common tangents to two circles of unequal radii.**

Let the radii of the circles with centres at O and O’ be r and r’, where r > r’, i.e., the radii are unequal.

We have to construct a direct common tangent to these circles.

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**Method of construction:**

- Let us draw two circles with centres at O and O’, the radii of which are r and r’ respectively, where r > r’.
- Let us join O and O’. Let OO’ = R, where R > r + r’.
- Let us now draw a third circle with centre at O and radius equal to (r – r’).
- Let us draw a tangent OA from O’ to this thiai circle.
- Let us join O and A and let us extend OA. Let extended OA intersects the greater circle with centre O at B.
- Let us draw the radius O’C on the same side of O’A and parallel to OB.
- Let us join B and C to get the line segment BC which is further extended to both the sides. Let the extended BC is PQ.

Hence PQ is the required direct common tangent to the circles with centres at O and O’ respectively.

**Proof:** O’A is a tangent to the circle with centre at O’ [as per construction] and OA is the radius through point of contact,

∴ OA ⊥ O’A, ∴ ∠O’AO = 90°.

Again, O’C || OB and BC is their transversal.

∴ ∠OBC = ∠O’CQ [∵ similar angles]

Again, ABCO’ is a parallelogram.

Since AB = OB – OA = r – (r — r’) = r – r + r’ = r’ = O’C and AB || O’C

∵ BC || AO’ and OB is their transversal.

∴ ∠OAO’ = ∠OBC [∵ similar angles]

∴ ∠OBC = 90°, [∵ ∠OAO’ = 90°]

∴ BC is a tangent to the circle with centre at O at the point B and BC is also a tangent to the circle with centre at O’ at the point C.

If BC is extended to both the sides, we get a line segment PQ (let).

Hence PQ is a direct common tangent to the circles. [Proved]

How constructions of a direct common tangent of two- circles of equal radii are made is discussed in the following construction.

**Construction of a direct common tangent to two circles of equal radii.**

Let two circles with centres at A and B are of equal radii. We have to construct a direct common tangent to the circles.

- Let us join A and B.
- Let us draw a perpendicular AC at A of the line segment AB. Let AC intersects the circle with centre at A at the point P.
- Let us draw an arc with centre at P and radius equal to AB. Let this arc intersects the circle with centre at B at a point Q.
- Let us join P and Q and let PQ is extended to both the sides.

Hence PQ is the required direct common tangent to the two circles.

**Construction of transverse common tangent to two circles:**

In the following construction the method of construction of drawing a transverse common tangent to two unequal circles is discussed.

**Construction of a transverse common tangent of two circles of unequal radii.**

Let the radii of two circles with centres at A and B be R and r (R > r) respectively.

We have to construct a transverse common tangent to both circles.

**Method of construction:**

- Let us draw two circles with centres at A and B, the radii of which are R and r respectively.
- Let us draw a third circle with centre at A and radius equal to the sum of the radii of the two given circles, i.e, equal to (R + r).
- Let us draw a tangent BP to this third circle from the point B which intersects the third circle at the point P.
- Let us join A and P. Let AP intersects the circle with centre at A at the point Q.
- Let us draw a line segment BD from the point B parallel to AQ and in its opposite direction. Let this line segment intersects the circle with centre at B at a point R.
- Let us join Q and R and let QR be extended to both sides.

Hence QR is the required transverse common tangent to two circles of unequal radii.

In the following construction, the method of construction of a transverse common tangent to two circles when the circles are of equal radii, have discussed.

**Class 10 Maths Wbbse Solutions**

**Construction of a transverse common tangent Of two circles of equal radii.**

Let the radii of two circles with centres at A and B arc equal (here r unit).

We have to construct a transverse common tangent to these two circles.

**Method of construction:**

- Let us construct two circles with centres at A and B with radius r.
- Let us join A and B.
- Let us draw the perpendicular bisector GH of AB, which intersects AB at a point P.
- Let us again draw the perpendicular bisector MN of AP. Let MN intersects AP at a point R.
- Let us draw an arc with centre at R and radius equal to RA. Let the arc intersect the circle with centre at A at a point C.
- Let us join C and P and let CP is extended to both sides upto E and F to get a line segment EPF which touches the circle with centre B at a point D.

Hence EPF is the required transverse common tangent to the circles of equal radii.

## Solid Geometry Chapter 8 Construction Of Tangent To A Circle Construction Of Common Tangents Examples

**Example 1. Construct two circles of radii 2 cm and 4 cm, the distance of whose centres is 8 cm. Construct a direct common tangent to these two circles**

Solution:

**Given:**

Two circles of radii 2 cm and 4 cm the distance of whose centres is 8 cm

Here, BC is the required direct common tangent to the circles with centres at O and O’ respectively.

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**Example 2. Construct two circles of radii 2 cm each, the distance of whose centres is 10 cm. Then construct a direct common tangent to two circles.**

Solution:

**Given:**

Two circles of radii 2 cm each, the distance of whose centres is 10 cm

Here PQ is the required direct common tangent to two circles with centres at O and O’.

**Example 3. Construct two circles of radii 2.5 cm each, the distance of whose centres is 8 cm. Then construct a transverse common tangent to these two circles.**

Solution:

**Given:**

Two circles of radii 2.5 cm each, the distance of whose centres is 8 cm.

Here CD is the required transverse common tangent to the circles with centres O and O’.

**Example 4. Construct two circles of radii 2 cm and 3 cm, the distance between whose centres is 8.7 cm. Then draw a transverse common tangent to these two circles.**

Solution:

Two circles of radii 2 cm and 3 cm, the distance between whose centres is 8.7 cm.

Here, PQ is the required transverse common tangent to the circles with centres at O’ and O.