MHT CET202416 May 2024Evening ShiftMathematicsCircleActual
One end of the diameter of the circle x^2+y^2-6 x-5 y-1=0 is (-1,3) , then the equation of the tangent at the other end of the diameter is
Options
- A8 x+y-58=0
- B8 x-2 y-52=0
- C8 x-y-54=0
- D8 x+2 y-60=0
Correct answer
C. 8 x-y-54=0
Step-by-step solution
If (x₁, y₁ ) is one end of a diameter of the circle x^2+y^2+2 ~g x+2 f y+ c =0 , then the other end is [- (x₁+2 ~g ),- (y₁+2 f ) ] The other end of x^2+y^2-6 x-5 y-1=0 is [-(-1-6),-(3-5)] i.e. (7,2) Equation of tangent at (7,2) is aligned & 7 x+2 y-3(x+7)- 5 2 (y+2)-1=0 & 8 x-y-54=0 aligned