Find the equation of all the line having slope equal to 3 and being tangent to the curve ` x^(2)+y^(2)=4 `

Explanation

Curve is ` x^(2)+y^(2)=4 ` ......(i)
Slope of tangent at any point P(x,y) to the curve is given by
`2x+2y""(dy)/(dx)=0`
`implies (dy)/(dx)=(-(x)/(y))=3 `
`implies -3y=x`
Using (i)
`9y^(2)+y^(2)=4 `
`implies 10y^(2)=4 `
`implies y= pm (2)/(sqrt(10))`
`x^(2)+(4)/(10)=4 `
`implies x^(2)=4-(4)/(10)=(36)/(10)`
`implies x= pm (6)/(sqrt(10))`
Equation of tangent at any point P(x,y) is given by `y-y_(0)=f'(x_(0))(x-x_(0))`
Here `(x_(0),y_(0))` is the point of contact
In given question slope =3 , so point of contacts will be
`((6)/(sqrt(10)),-(2)/(sqrt(10))) and (-(6)/(sqrt(10)),(2)/(sqrt(10)))`
There will be two tangent having slope 3 and equation will be
(i) `y+(2)/(sqrt(10))=3(x-(6)/(sqrt(10)))`
`implies 3x-y=(20)/(sqrt(10)) or 3x-y=2sqrt(10)`
(ii) ` y-(2)/(sqrt(10))=3(x+(6)/(sqrt(10)))`
`implies 3x-y=-2sqrt(10)`