In this page equation of the tangent we are going to see how to find the equation of a tangent to the circle.First let us see the definition.

A tangent to a circle is a straight line which intersects (touches) the circle exactly one point.We can draw only two tangents to a circle from the point outside to a circle.

**To get the equation of a tangent at(x₁,y₁) we have to replace x²
by xx₁, y² by yy₁ , x as (x + x₁)/2 and y as (y + y₁)/2 in the equation of a circle.**

Here we are going to see the examples to find the equation of the tangent to the circle.

__Example 1:__

Find the equation of a tangent to the circle x² + y² = 25 at (4,3).

**Solution:**

As per the procedure we have to replace x² by xx₁ and y² by yy₁.

Equation of a circle x² + y² = 25

Equation of the tangent

xx₁ + yy₁ = 25

Here (x₁,y₁) is (4,3)

x(4) + y(3) = 25

** 4x + 3y = 25**

__Example 2:__

Find the equation of tangent to the circle x² + y² - 4x -3y +12 = 0 at (2,3).

__Solution:__

Equation of the circle x² + y² - 4x -6y +12 = 0

We have to replace x² by xx₁ y² by yy₁,x as (x+x₁)/2 and y as (y +y₁)/2

xx₁ + yy₁ - 4 {(x+x₁)/2} - 6 {(y +y₁)/2} + 12 = 0

Equation of a circle at a point (2,3)

x(2) + y(3) + 2(x +2) -3(y+ 3) + 12 = 0

2x + 3y + 2x + 4 - 3y - 9 + 12 = 0

4x + 16 - 9 = 0

** 4x + 7 = 0**

__Example 3:__

Find the equation of the tangent to x² + y² - 4x -4y -8 = 0 at (-2,-2).

**Solution:**

Equation of the circle x² + y² - 4x -4y -8 = 0

We have to replace x² by xx₁ y² by yy₁,x as (x+x₁)/2 and y as (y +y₁)/2

xx₁ + yy₁ - 4 {(x+x₁)/2} - 4 {(y +y₁)/2} - 8 = 0

Equation of a circle at a point (-2,-2)

x(-2) + y(-2) + 2(x -2) -2(y - 2) -8 = 0

-2x - 2y + 2x - 4 - 2y + 4 - 8 = 0

-4y - 8 = 0

4y + 8 = 0

** y + 2 = 0**

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