## LENGTH OF THE TANGENT

Length of the tangent :

Here we are going to see how to find the length of the tangent to the circle from the point (x₁,y₁)

Length of the tangent = √ (x₁² + y₁² + 2gx₁ +2fy₁ + c)

(i) If the length is 0.Then we can say the given point must be on the circle.

(ii) If the length is > 0 .Then we can say the point must be outside the circle.

(iii) If the length is < 0 .Then we can say the point must be inside the circle.

Example 1 :

Find the length of tangent to the circle x² + y² -4x - 3y + 12 = 0from the point (2,3)

Solution :

Length of the tangent  =  √ (x₁² + y₁² + 2gx₁ +2fy₁ + c)

Here x₁ = 2 and y₁ = 3

=  √ (2)² + 3² - 4(2) - 3(3) + 12

=  √ 4 + 9 - 8 - 9 + 12

=  √4 + 12 - 8

=  √ 16 - 8

=  √8

=  2√2 units

Example 2 :

Show that the point (2,-1) lies out side the circle x²+y²-6x-8y+12 = 0

Solution :

Length of the tangent

= √ (x₁² + y₁² + 2gx₁ +2fy₁ + c)

Here x₁ = 2 and y₁ = -1

=  √ (2)² + (-1)² - 6(2) - 8 (-1) + 12

=  √ 4 + 1 - 16 + 8 + 12

=  √5 + 8 + 12 + 1 - 8

=  √ 26 - 8

=  √18 =  3√2 units

The length of tangent is positive so the given point lies outside the circle.

Example 3 :

Find the length of tangent to the circle  x²+y²-4x+8y-5 = 0 from the point (2,1)

Solution :

Length of the tangent = √ (x₁² + y₁² + 2gx₁ +2fy₁ + c)

Here x₁ = 2 and y₁ = 1

=  √ 2² + 1² - 4(2) + 8(1) -5

=  √ 4 + 1 - 8 + 8 -5

=  √(5 - 5)

=  √0  =  0

The length of tangent is zero so the given point lies on the circle.length of the tangent

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