Equation of a common tangent with positive slope to the circle x^ 2 +y^ 2 -8 x=0 as well…
Mathematics · JEE Main · NTA Exams — Co-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,
- \(2 x+\sqrt{5} y+4=0\)
- \(2 x-\sqrt{5} y+4=0\)
- \(3 x-4 y+8=0\)
- \(4 x-3 y+4=0\)
Answer
(B) 2 x- 5 y+4=0
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