## 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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