Tangent:
If a line touches the given circle at only one point, then it is called 'tangent' to the circle.
Important!
The common point where the circle and the tangent intersect is called the point of contact.
Theorem on circles and tangents:
Statement:
The tangent at any point of a circle is perpendicular to the radius through the point of contact.
Proof:
Consider a circle with centre \(O\).
Let \(AB\) be the tangent to the circle at the point \(P\).
To prove:
The line \(OP\) is perpendicular to \(AB\).
Proof:
Take a point \(Q\) other than \(P\) on the tangent \(AB\) and join \(OQ\).
Here, \(Q\) must lie outside the circle.
Thus, \(OQ\) is longer than \(OP\).
That is \(OQ>OP\) at every point on \(AB\) except at \(P\).
Therefore, the point \(P\) is at the shortest distance from the centre \(O\).
We know that:
Out of all the line segments drawn from the point to points of a line not passing through the point, the smallest is the perpendicular to the line.
By the theorem, \(OP\) is perpendicular to \(AB\).
Hence, the proof.
Example:
In the above given figure, if \(OP = 3\ cm\) and \(PQ = 4\ cm\), find the length of \(OQ\).
Solution:
By the result, \(\angle OPQ = 90^\circ\)
So, \(OPQ\) is a right-angled triangle.
By the Pythagoras theorem, we have:
In a right triangle, the square of the hypotenuse is equal to the sum of the squares of the other two sides.
\(OQ^2 = OP^2 + PQ^2\)
\(OQ^2 = 3^2 + 4^2\)
\(OQ^2 = 9+16\)
\(OQ^2 = 25\)
\(OQ = \sqrt{25}\)
\(OQ = 5\)
Therefore, the measure of \(OQ = 5\ cm\).
Important!
- There are no tangent can be drawn from an interior point of the circle.
- Only one tangent can be drawn at any point on a circle.
- Two tangents can be drawn from any exterior point of a circle.
Theorem \(2\) on circles and tangents
Statement:
The lengths of tangents drawn from an exterior point to a circle are equal.
Proof:
Consider a circle with centre \(O\).
Let \(PA\) and \(PB\) be the two tangents drawn from the external point \(P\) to the circle.
Construction:
Join \(OA\), \(OB\) and \(OP\).
To prove:
The tangent \(PA=\) The tangent \(PB\).
Proof:
By theorem \(1\), we have:
The tangent at any point of a circle is perpendicualr to the radius through the point of contact.
\(OB\perp PB\) and \(OA\perp PA\)
Here, \(OA\) and \(OB\) are radius. Hence, they are equal.
The side \(OP\) is a common side to the triangles \(AOP\) and \(BOP\).
Therefore, by \(RHS\) rule, if the length of the hypotenuse and one side of one triangle is equal to the length of the hypotenuse and corresponding side of the other triangle, then the two triangles are congruent.
The triangles \(AOP\) and \(BOP\) are congruent.
We know that the corresponding parts of the congruent triangles are equal.
Therefore, \(PA=PB\).
Example:
In the above figure if \(OB=3\ cm\) and \(OP=5\ cm\), find the length of \(PA\).
Solution:
By the theorem \(1\), we have:
A tangent at any point on a circle and the radius through the point are perpendicular to each other.
\(\angle OBP=90^\circ\).
So, \(OBP\) is a right angled triangle.
By the Pythagoras theorem, we have:
In a right triangle, the square of the hypotenuse is equal to the sum of the squares of the other two sides.
\(OP^2 = OB^2 + PB^2\)
\(PB^2 = OP^2-OB^2\)
\(PB^2 = 5^2-3^2\)
\(PB^2 = 16\)
\(\Rightarrow PB=\sqrt{16}\)
\(PB = 4\)
Thus, the measure of \(PB = 4\ cm\).
By theorem \(2\), we have:
The length of the two tangents drawn from an exterior point to a circle are equal.
Hence \(PA = PB\)
Therefore, the measure of \(PA = 4\ cm\).