Question

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

• A. \(\sqrt{3}y = 3x+1\)
• B. \(\sqrt{3}y = -(x+3)\)
• C. \(\sqrt{3}y = x+3\)
• D. \(\sqrt{3}y = -(3x+1)\)

Hint:

Use y = mx + 1/m as tangent to the parabola, then the distance from the circle centre must equal the radius.

Answer 1

The Correct Option is C

Step 1: Tangent form
Any tangent of $y^2 = 4x$ is $y = mx + 1/m$.

Step 2: Circle condition
Distance from $(3,0)$ to the line must be 3. Squaring gives $3m^2 = 1$, so $m = \pm 1/\sqrt3$.

Step 3: Sign
The contact point $(1/m^2, 2/m) = (3, 2\sqrt3)$ for $m > 0$ is above the axis. The line is $y = x/\sqrt3 + \sqrt3$, option (C).

Final Answer:
Option (C). \[ \boxed{\sqrt{3}\,y = x + 3} \]