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find the equation of tangent to the circle @@x^2+y^2@@-80x-60y+2100=0 at the point nearest to origin.

find the equation of tangent to the circle @@x^2+y^2@@-80x-60y+2100=0 at the point nearest to origin.

Grade:10

2 Answers

Chetan Mandayam Nayakar
312 Points
10 years ago

the equation of the circle can be written as

(ax-(40/a))2 + (ay-(30/a))2 =(2500/a^2) -2100  equation(1)

centre is (40/a^2,30/a^2). thus centre lies on the line y=3x/4...eqn(2)

the coordinates of the pont where the circle intersects this line are given by solving the simultaneous equations (1) and (2).

we must substitute 'y' by 3x/4

we get x=(40/a2)(1±√(1-21a2)), from the two answers, we obviously have to chose the one with lower value value of x,

thus c =(3/4)(40/a2)(1-√(1-21a2)) + (40/a2)(1-√(1-21a2))=(70/a2)(1-√(1-21a2)),

therefore, answer is y=-(4x/3) + (70/a2)(1-√(1-21a2)),

 

Shivansh Sahai
9 Points
2 years ago
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