FIND THE EQUATION OF THE TANGENT TO THE CIRCLE AT THE POINT

About "Find the equation of the tangent to the circle at the point"

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

x2  =  xx1, y2  =  yy1, x = (x + x1)/2, y  =  (y + y1)/2

Equation of tangent at the point (x1, y1) to the circle

xx1 + yy1 + g(x + x1) + f(y + y1) + c  =  0

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

Example 1 :

Find the equation of the tangent to x2 + y2 − 2x − 10y + 1 = 0 at (− 3, 2)

Solution :

Equation of tangent at (x1, y1) :

xx1 + yy− 2((x + x1)/2) − 10((y + y1)/2) + 1 = 0 

xx1 + yy− (x + x1) − 5(y + y1)  + 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 x2 + y2 − 4x + 2y − 21 = 0 at (1, 4)

Solution :

Equation of tangent at (x1, y1) :

xx1 + yy1 - 4((x + x1)/2) + 2((y + y1)/2) - 21  =  0 

xx1 + yy− 2(x + x1) + (y + y1)  - 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 x2 + y2 = 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 + √(1 + m2)

here "m" stands for slope of the tangent,

Since the tangent line drawn to the circle x2 + y2 = 16 is perpendicular to the line x + y = 8, the product of slopes will be equal to -1.

m  =  -1/(-1)  ==>  1

y = mx + √(1 + m2)

y  =  1x + 4 √(1 + 12)

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 x2 + y2 = 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

x2 + y2 = 16 

a2  =  16 ==> a = 4

y = mx + √(1 + m2)

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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