Equation of a common tangent with positive slope to the circle x^ 2 +y^ 2 -8 x=0 as well…

Mathematics · JEE Main · NTA ExamsCo-ordinate Geometry

Equation of a common tangent with positive slope to the circle \(x^{2}+y^{2}-8 x=0\) as well as to the hyperbola \(\frac{x^{2}}{9}-\frac{y^{2}}{4}=1\) is,
  1. \(2 x+\sqrt{5} y+4=0\)
  2. \(2 x-\sqrt{5} y+4=0\)
  3. \(3 x-4 y+8=0\)
  4. \(4 x-3 y+4=0\)

Answer

(B) 2 x- 5 y+4=0

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