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Answer

`y=pm3sqrt((2)/(7))xpm(15)/(sqrt7)`

Explanation

Given equation of hyperbola is
`9x^(2)-16y^(2)=144`
`"or "(x^(2))/(16)-(y^(2))/(9)=1`
The equation of tangent to the hyperbola having slope m is brgt `y=mxpmsqrt(16m^(2)-9)`
If it touches the circle, then the distance of the line from the centre of the circle is the radius of the circle. Hence,
`(sqrt(16m^(2)-9))/(sqrt(m^(2)+1))=3`
`"or "9m^(2)+9=16m^(2)-9`
`"or "7m^(2)=18`
`"or "m=pm3sqrt((2)/(7))`

So, the equation of tangent is
`y=pm3sqrt((2)/(7))xpm(15)/(sqrt7)`