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Grade: 12th pass


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

3 years ago

Answers : (1)

24742 Points

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


3 years ago
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