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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
Line parallel to 3x-4y+6=0 is 3x-4y=cLet c be any constant. A tangent have only one common point with circle. Perpendicular distance of tangent from center of circle is equal radius. 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= 3So 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 be3x-4y=10 and3x-4y=-20
Line parallel to 3x-4y+6=0 is 3x-4y=c
Let c be any constant.
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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