### Theory:

We learned about circles and tangents in previous chapters. The number of tangents for the circle will vary depending on the point located.
(i) If a point lies inside the circle, then there cannot be a tangent passing through that point.

(ii) If a point lies on the circle, only one tangent can be possible through that point, and it is perpendicular to the radius.

(iii) If a point lies outside the circle, then two tangents can be possible through that point.
Now, we will see how to construct a pair of tangents to a circle from an external point.
Example:
1. Draw a circle of diameter $$6$$ $$cm$$. From a point $$P$$, which is $$8$$ $$cm$$ away from its centre, construct two tangents $$PA$$ and $$PB$$ to the circle and measure their lengths.

Solution:

The diameter of the circle $$=$$ $$6$$ $$cm$$.

The radius of the circle $$=$$ $$\frac{Diameter}{2}$$

Radius $$=$$ $$\frac{6}{2}$$ $$=$$ $$3$$ $$cm$$ Steps of construction:

Step 1: With $$O$$ as the centre, draw a circle of radius $$3$$ $$cm$$.

Step 2: Draw a line $$OP$$ of length $$8$$ $$cm$$.

Step 3: Draw a perpendicular bisector of $$OP$$, which cuts $$OP$$ at $$M$$.

Step 4: With $$M$$ as centre and $$MO$$ as radius, draw a circle that cuts the previous circle at $$A$$ and $$B$$.

Step 5: Join $$AP$$ and $$BP$$. $$AP$$ and $$BP$$ are the required tangents. Thus the length of the tangents are $$PA = PB = 7.4$$ $$cm$$.

JUSTIFICATION:

In the right angle triangle $$OAP$$ by the Pythagoras theorem, we have:

$$OP^2$$ $$=$$ $$OA^2$$ $$+$$ $$PA^2$$

$$\Rightarrow$$ $$PA^2$$ $$=$$ $$OP^2$$ $$-$$ $$OA^2$$

$$\Rightarrow$$ $$PA^2$$ $$=$$ $$8^2$$ $$-$$ $$3^2$$

$$\Rightarrow$$ $$PA^2$$ $$=$$ $$64 – 9$$

$$\Rightarrow$$ $$PA^2$$ $$=$$ $$55$$

$$\Rightarrow PA = \sqrt{55}$$

$$\Rightarrow$$ $$PA$$ $$=$$ $$7.4$$ $$cm$$ (approximately).

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

$$\Rightarrow$$ $$PA$$ $$=$$ $$PB$$ $$=$$ $$7.4$$ $$cm$$

2. Draw a pair of tangents to a circle of radius $$3.5 \ cm$$, which are inclined to each other at an angle of $$60^\circ$$.

Solution:

The radius of the circle $$=$$ $$3.5 \ cm$$

The angle between tangents $$=$$ $$60^\circ$$

The angle between the two tangents drawn from an external point to a circle is supplementary to the angle subtended by the line segment joining the points of contact at the centre.

That is, the angle between tangents $$+$$ the angle subtended at the centre $$=$$ $$180^\circ$$

$$60^\circ$$ $$+$$ the angle subtended at the centre $$=$$ $$180^\circ$$

The angle subtended at the centre $$=$$ $$180^\circ$$ $$-$$ $$60^\circ$$

The angle subtended at the centre $$=$$ $$120^\circ$$ Steps of construction:

Step 1: Draw a circle of radius $$3.5 \ cm$$.

Step 2: Draw a horizontal radius $$OA$$.

Step 3: Construct an angle $$120^\circ$$ at the centre $$O$$. Let the ray of an angle intersect the circle at point $$B$$.

Step 4: We know that, the tangent is perpendicular to the radius. As a result, construct $$90^\circ$$ from points $$A$$ and $$B$$.

Step 5: Let the two perpendicular rays intersect and mark the point of intersection as $$P$$. Thus, $$PA$$ and $$PB$$ are tangents at an angle of $$60^\circ$$.

JUSTIFICATION:

We need to justify that $$PA$$ and $$PB$$ are tangents to a circle at an angle of $$60^\circ$$.

$$OA$$ and $$OB$$ are radii.

By construction, $$\angle PAO = 90^\circ$$.

$$PA \perp OA$$

A tangent at any point on a circle and the radius through the point are perpendicular.

Thus, $$PA$$ is tangent to the circle.

Similarly, by construction, $$\angle PBO = 90^\circ$$.

$$PB \perp OB$$

A tangent at any point on a circle and the radius through the point are perpendicular.

Thus, $$PB$$ is tangent to the circle.