**Find the equation of the tangent to the circle at the point :**

Here we are going to see how to find equation of the tangent to the circle at the given point.

**What is tangent to the circle ?**

A tangent to a circle is a straight line which intersects (touches) the circle in exactly one point.

To find the equation of tangent at the given point, we have to replace the following

x^{2} = xx_{1}, y^{2} = yy_{1}, x = (x + x_{1})/2, y = (y + y_{1})/2

**xx _{1} + yy_{1} + g(x + x_{1}) + f(y + y_{1}) + c = 0**

Let us look into some examples to understand the above concept.

**Example 1 :**

Find the equation of the tangent to x^{2} + y^{2} − 2x − 10y + 1 = 0 at (− 3, 2)

**Solution :**

Equation of tangent at (x_{1}, y_{1}) :

xx_{1} + yy_{1 }− 2((x + x_{1})/2) − 10((y + y_{1})/2) + 1 = 0

xx_{1} + yy_{1 }− (x + x_{1}) − 5(y + y_{1}) + 1 = 0

at the point (-3, 2)

x(-3) + y(2)_{ }− (x - 3) − 5(y + 2) + 1 = 0

-3x + 2y - x + 3 - 5y - 10 + 1 = 0

-4x - 3y - 6 = 0

Multiply by (-) on both sides

4x + 3y + 6 = 0

Let us look into the next example on "Find the equation of the tangent to the circle at the point".

**Example 2 :**

Find the equation of the tangent to the circle x^{2} + y^{2} − 4x + 2y − 21 = 0 at (1, 4)

**Solution :**

Equation of tangent at (x_{1}, y_{1}) :

xx_{1} + yy_{1 }- 4((x + x_{1})/2) + 2((y + y_{1})/2) - 21 = 0

xx_{1} + yy_{1 }− 2(x + x_{1}) + (y + y_{1}) - 21 = 0

at the point (1, 4)

x(1) + y(4)_{ }− 2(x + 1) + (y + 4) - 21 = 0

x + 4y - 2x - 2 + y + 4 - 21 = 0

-x + 5y - 23 + 4 = 0

x - 5y - 19 = 0

**Example 3 :**

Find the equation of the tangent to the circle x^{2} + y^{2} = 16 which are

(i) perpendicular and

(ii) parallel to the line x + y = 8

**Solution :**

Equation of tangent to the circle will be in the form

y = mx + a √(1 + m^{2})

here "m" stands for slope of the tangent,

Since the tangent line drawn to the circle x^{2} + y^{2} = 16 is perpendicular to the line x + y = 8, the product of slopes will be equal to -1.

m = -1/(-1) ==> 1

y = mx + a √(1 + m^{2})

y = 1x + 4 √(1 + 1^{2})

y = x + 4 √2

x - y + 4 √2 = 0

Hence the equation of the tangent perpendicular to the given line is x - y + 4 √2 = 0.

(ii) Since the tangent line drawn to the circle x^{2} + y^{2} = 16 is parallel to the line x + y = 8, the slopes of the tangent line and given line will be equal.

m = -1/1 = -1

x^{2} + y^{2} = 16

a^{2} = 16 ==> a = 4

y = mx + a √(1 + m^{2})

y = -1x + 4 √(1 + (-1)^{2})

y = -x + 4 √2

x + y - 4 √2 = 0

Hence the equation of the tangent parallel to the given line is x + y - 4 √2 = 0.

After having gone through the stuff given above, we hope that the students would have understood "Find the equation of the tangent to the circle at the point".

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