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Find the equation of the tangent of the circle x^2+y^2-2x-4y-4=0 which is perpendicular to the line 3x-4y+6=0

Find the equation of the tangent of the circle x^2+y^2-2x-4y-4=0 which is perpendicular to the line 3x-4y+6=0

Grade:12th pass

1 Answers

Arun
25763 Points
4 years ago

Line parallel to 3x-4y+6=0 is 3x-4y=c

Let c be any constant.

  1. A tangent have only one common point with circle.
  2. Perpendicular distance of tangent from center of circle is equal radius.
  3. Two tangent can be parallel.

Here equation of circle will be

(x-1)^2 +(y-2)^2 = 3^2.

So center of circle is (1,2)

Radius= 3

So distance of 3x-4y-c= 0 from (1,2) is equal to 3.

So,| (3–4*2-c)| /(3^2+4^2)^0.5 =3.

=> so c= -20 ,10.

So eqns will be

3x-4y=10 and

3x-4y=-20

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