Result 3 on circles and tangents — lesson. Mathematics State Board, Class 10.

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Result $3$ :

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

Explanation:

The lengths of the two tangents

$PA$ and

$PB$ drawn from an exterior point

$P$ to a circle are equal.

$\Rightarrow$

$PA = PB$ .

Proof of the result:

By the result

$1$ , we have:

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

$OB$

$\perp$

$PB$ 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

$BOP$ .

Thus, the triangles

$AOP$ and

$BOP$ are congruent.

Therefore,

$PA = PB$ .

Example:

In the above given figure if

$OB$

$=$

$3$

$cm$ and

$OP$

$=$

$5$

$cm$ , find the length of

$PA$ .

Solution:

By the result

$1$ , we have:

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

$\angle OPB$

$=$

$90^{\circ}$ .

So,

$OPB$ 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.

$OP^2$

$=$

$OB^2$

$+$

$PB^2$

$=$

$OP^2$

$-$

$OB^2$

$PB^2$

$=$

$5^2$

$-$

$3^2$

$PB^2$

$=$

$25 - 9$

$PB^2$

$=$

$16$

$\Rightarrow$

$PB$

$=$

$\sqrt{16}$

$PB$

$=$

$4$

Thus, the measure of

$PB$

$=$

$4$

$cm$

By the result

$3$ , we have:

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

Hence,

$PA$

$=$

$PB$ .

Therefore, the measure of

$PA$

$=$

$4$

$cm$

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4. Result 3 on circles and tangents

Theory: