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

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)),