Let the locus of the point of intersection of the perpendicular tangents drawn to the circle x^2+y^2+6 x-4 y-12=0 be the circle S . Then the equation of the tangent drawn to S which is perpendicular to the line 6 x -4 y + k =0 is

Step-by-step solution

Given circle is x^2+y^2+6 x-4 y-12=0 Centre =(-3,2) ; radius =5 Centre of circle ( S )=(-3,2) Since A P B=90^ , therefore O A P B is square and O A=5 Radius of circle (S)=O P=5 2 Now, equation of circle S is (x+3)^2+(y-2)^2=50 (i) For slope of tangent to S : 2(x+3)+2(y-2) d y d x =0 m₁=- (x+3) y-2 Slope of given lines is (m₂ )= 3 2 Since both are perpendicular aligned & -(x+3) y-2 3 2 =-1 x+3= 2 3 (y-2) & 4 9 (y-2)^2+(y-2)^2=50 [From equation (i)] & y= 15 2 13 +2 x= 30 2 39 -3 aligned Now equation of tangent is ali