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: Understanding the Concept
For \(y^2 = 4x\), a tangent with slope \(m\) is \(y = mx + \frac1m\). The same line must touch the circle with centre \((3, 0)\) and radius 3, so its distance from \((3, 0)\) is 3.

Step 2: Apply the condition
Write the tangent as \(mx - y + \frac1m = 0\). Distance from \((3,0)\):
\[ \frac{|3m + \frac1m|}{\sqrt{m^2 + 1}} = 3 \Rightarrow \left(3m + \frac1m\right)^2 = 9(m^2 + 1) \]
\[ 9m^2 + 6 + \frac{1}{m^2} = 9m^2 + 9 \Rightarrow \frac{1}{m^2} = 3 \Rightarrow m = \pm\frac{1}{\sqrt3} \]

Step 3: Choose the one above the X-axis
The point of contact on the parabola is \(\left(\frac{1}{m^2}, \frac2m\right)\). It lies above the X-axis when \(m > 0\). So \(m = \frac{1}{\sqrt3}\) and \(y = \frac{x}{\sqrt3} + \sqrt3\), that is \(\sqrt3\,y = x + 3\).
Option (B), with negative slope, touches below the axis. Options (A) and (D) fail the distance condition.

Final Answer:
The tangent is \(\sqrt3 y = x + 3\), option (C). \[ \boxed{\sqrt{3}\,y = x + 3} \]