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
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 + yy1 − 2((x + x1)/2) − 10((y + y1)/2) + 1 = 0
xx1 + yy1 − (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 + yy1 − 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 + a √(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 + a √(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 + a √(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.
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