PDF chapter test TRY NOW

*Statement*:

The tangent at any point of a circle is perpendicular to the radius through the point of contact.

*Proof for the theorem*:

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 a 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.

\(OQ^2\) \(=\) \(OP^2\) \(+\) \(PQ^2\)

\(OQ^2\) \(=\) \(3^2\) \(+\) \(4^2\)

\(OQ^2\) \(=\) \(9 + 16\)

\(OQ^2\) \(=\) \(25\)

\(\Rightarrow\) \(OQ\) \(=\) \(\sqrt{25}\)

\(OQ\) \(=\) \(5\)

Therefore, the measure of \(OQ\) \(=\) \(5\) \(cm\)

Important!

- By this theorem, we can say that at any point on the circle, there can be one and only one tangent.
- The line containing the radius through the point of contact is also called the normal to the circle at the point.