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`

`2x-sqrt5y-20=0`

`2x-sqrt5y+4=0`

`3x-4y+8=0`

`4x-3y+4=0`

Text Solution

Verified by Experts

The correct Answer is:

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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The circle x ^(2) +y^(2) -8 x+4=0 touches -

x-axis

y-axis

both axes

neither x-axis nor y-axis

The circle x^2 + y^2 - 8x + 4y +4 = 0 touches

x-axis

y-axis

both axis

Neither x-axis nor y-axis

The circle x^2 + y^2 - 2x - 4y + 1 = 0 and x^2 + y^2 + 4x + 4y - 1 = 0

touches internally

touch externally

have 3x + 4y - 1 = 0 the common tangent at the point of contact

have 3x + 4y + 1 = 0 as the common tangent at the point of contact