Question

Equation of a common tangent to the circle x² + y² – 6x = 0 and the parabola y² = 4x is​

Answer 1

$\large\underline\mathrm\red{Solution}$

• Any tangent to y² = 4x is of the form y = Mx + 1/m, ( a = 1 )
• this touches the circle ( x - 3 )² + y² = 9.

$\implies$|(m(3) + 1/m - 0) / (√m² + 1)| = 3

$\large\mathrm\red{Centre \ of \: the \: circle \: is \: (3, \: 0) \: and \: radius \: is \: 3.}$

$\implies$3m² + 1 / m = +, - 3√m² + 1

$\implies$3m² + 1 = +, - 3m√m² + 1

$\implies$9m⁴ + 1 + 6m² = 9m⁴ + 9m²

$\implies$3m² = 1

$\implies$m = +, - 1/√3

$\large\mathrm\red{Thus}$, the tangent touches the parabola and circle above the x - axis, then slope m should be positive.

m = 1/√3 and the equation is y = 1/√3 x + √3

√3y = x + 3.

__________________________________

Answer 2

$\large\underline\mathrm\red{Solution}$

Any tangent to y² = 4x is of the form y = Mx + 1/m, ( a = 1 )

this touches the circle ( x - 3 )² + y² = 9.

⟹ |(m(3) + 1/m - 0) / (√m² + 1)| = 3

⟹ 3m² + 1 / m = +, - 3√m² + 1

⟹ 3m² + 1 = +, - 3m√m² + 1

⟹ 9m⁴ + 1 + 6m² = 9m⁴ + 9m²

⟹ 3m² = 1

⟹ m = +, - 1/√3

$\large\mathrm\red{Thus}$ , the tangent touches the parabola and circle above the x - axis, then slope m should be positive.

m = 1/√3 and the equation is y = 1/√3 x + √3

√3y = x + 3.

__________________________________