Pair of tangents
Pair of Tangents
When two tangents are drawn from an external point to a circle, they form a pair of tangents. These tangents have several interesting properties and are an important topic in geometry, especially in the context of circles.
Properties of a Pair of Tangents
- Equal Lengths: Tangents drawn from an external point to a circle are of equal lengths.
- Angle Between the Tangents: The angle between the tangents is equal to the angle subtended by the line segment, joining the center of the circle to the external point, at the center.
- Tangent-Radius Perpendicularity: The tangent to a circle is perpendicular to the radius at the point of contact.
Let's denote the following:
- ( P ): External point from which tangents ( PA ) and ( PB ) are drawn.
- ( O ): Center of the circle.
- ( A ) and ( B ): Points of contact of the tangents with the circle.
- ( r ): Radius of the circle.
- ( OP ): Distance from the external point to the center of the circle.
Table of Differences and Important Points
| Property | Description |
|---|---|
| Length of Tangents | ( PA = PB ) |
| Angle Between Tangents | ( \angle APB ) is equal to ( \angle AOB ) |
| Tangent-Radius Perpendicularity | ( OA \perp PA ) and ( OB \perp PB ) |
| Area of Quadrilateral | Area of quadrilateral ( OAPB ) can be found using the formula for cyclic quadrilaterals if ( OP ) is known. |
Formulas
The length of the tangents from an external point ( P ) to a circle with radius ( r ) and center ( O ) can be found using the Pythagorean theorem:
[ PA = PB = \sqrt{OP^2 - r^2} ]
The angle ( \theta ) between the tangents ( PA ) and ( PB ) can be found using the formula:
[ \tan(\theta) = \frac{2r}{OP - r} ]
Examples
Example 1: Finding the Length of Tangents
Given: A circle with radius ( r = 5 ) units and an external point ( P ) located ( 13 ) units away from the center of the circle.
Find: The length of the tangents from ( P ) to the circle.
Solution:
Using the formula for the length of the tangents:
[ PA = PB = \sqrt{OP^2 - r^2} = \sqrt{13^2 - 5^2} = \sqrt{169 - 25} = \sqrt{144} = 12 \text{ units} ]
Example 2: Finding the Angle Between Tangents
Given: A circle with radius ( r = 3 ) units and an external point ( P ) located ( 10 ) units away from the center of the circle.
Find: The angle between the tangents from ( P ) to the circle.
Solution:
Using the formula for the angle between the tangents:
[ \tan(\theta) = \frac{2r}{OP - r} = \frac{2 \cdot 3}{10 - 3} = \frac{6}{7} ]
To find the angle ( \theta ), we take the arctangent of ( \frac{6}{7} ):
[ \theta = \arctan\left(\frac{6}{7}\right) \approx 40.60^\circ ]
Example 3: Area of Quadrilateral ( OAPB )
Given: A circle with radius ( r = 4 ) units and an external point ( P ) located ( 8 ) units away from the center of the circle.
Find: The area of quadrilateral ( OAPB ).
Solution:
First, find the length of the tangents:
[ PA = PB = \sqrt{OP^2 - r^2} = \sqrt{8^2 - 4^2} = \sqrt{64 - 16} = \sqrt{48} = 4\sqrt{3} \text{ units} ]
Since ( OAPB ) is a cyclic quadrilateral, we can use Brahmagupta's formula or we can divide the quadrilateral into two triangles ( OAP ) and ( OBP ) and find the area of each using the formula for the area of a triangle:
[ \text{Area} = \frac{1}{2} \times \text{base} \times \text{height} ]
For triangle ( OAP ):
[ \text{Area}_{OAP} = \frac{1}{2} \times OP \times PA = \frac{1}{2} \times 8 \times 4\sqrt{3} = 16\sqrt{3} \text{ square units} ]
Since ( OAP ) and ( OBP ) are congruent, the area of quadrilateral ( OAPB ) is:
[ \text{Area}{OAPB} = 2 \times \text{Area}{OAP} = 2 \times 16\sqrt{3} = 32\sqrt{3} \text{ square units} ]
These examples illustrate the application of the properties and formulas related to the pair of tangents drawn from an external point to a circle. Understanding these concepts is crucial for solving problems in geometry, especially in exams that involve circle theorems and properties.