AP EAMCET2010MathematicsCircle
If the lengths of tangents drawn to the circles x^2+y^2-8 x+40=05 x^2+5 y^2-25 x+80=0x^2+y^2-8 x+16 y+160=0 From the point P are equal, then P is equal to
Options
- A(8, 15 2 )
- B(-8, 15 2 )
- C(8, -15 2 )
- D(-8, -15 2 )
Correct answer
C. (8, -15 2 )
Step-by-step solution
Let P (x₁, y₁ ) be the point from which the tangents are drawn to the circles S₁ x^2+y^2-8 x+40=0S₂ 5 x^2+5 y^2-25 x+80=0S₃ x^2+y^2-8 x+16 y+160=0 Since, the length of the tangent from P to the circle S₁, S₂, S₃ are equal S₁ = S₂ = S₃ S₁=S₂=S₃x₁^2+y₁^2-8 x₁+40=5 x₁^2+5 x₂^2-25 x₁+80=x₁^2+y^2-8 x₁+16 y₁+160 (i) Taking first and third part of above relation (i), we get -40+16 y₁+160=016 y₁+120=0y₁=- 120 16 - 15 2 Taking first and second part of the relation (i). -3 x₁+24=0x₁= 24 3 x₁=8 Hence, the point P is (8, -15 2