Question

The equation of the common tangent touching the parabola y^2=4x and
the circle (x-3)^2+y^2=9 above the x-axis 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.

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