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 \(=\) \(\frac{6}{2}\) \(=\) \(3\) \(cm\)
 
Cons3GIF.gif
 
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 \(=\) \(120^\circ\)
 
Circle.gif
 
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.