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The circle x^2+y^2-8x=0 and hyperbola x^...

The circle `x^2+y^2-8x=0` and hyperbola `x^2/9-y^2/4=1` I intersect at the points A and B. Equation of a common tangent with positive slope to the circle as well as to the hyperbola is

A

`2x-sqrt5y-20=0`

B

`2x-sqrt5y+4=0`

C

`3x-4y+8=0`

D

`4x-3y+4=0`

Text Solution

Verified by Experts

The correct Answer is:
B

A tangent to `(x^(2))/(9)-(y^(2))/(4)=1,` having slope m, is
`(x^(2))/(9)-(y^(2))/(4)=1`, having slope m, is
`y=mx+sqrt(9m^(2)-4),mgt0`
It is tangent to `x^(2)+y^(2)-8x=0`. Therefore, its distance from the centre of the circle is equal to the radius of circle.
`therefore" "(4x+sqrt(9m^(2)-4))/(sqrt(1+m^(2)))=4`
`"or "495m^(4)+104m^(2)-400=0`
`"or "m^(2)=(4)/(5)or m =(2)/(sqrt5)`
Therefore, the tangent is
`y=(2)/(sqrt5)x+(4)/(sqrt5)`
`"or "2x-sqrt5y+4=0`
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