 UPSKILL MATH PLUS

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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.