Question

A common tangent to the circle \( x^2 + y^2 = 9 \) and the parabola \( y^2 = 8x \) is 

• A. \( 3x - \sqrt{3}y + 2 = 0 \)
• B. \( x - \sqrt{3}y + 6 = 0 \)
• C. \( 2x - \sqrt{3}y + 3 = 0 \)
• D. \( x - 3y + 6 = 0 \)

Hint:

To find the common tangent between a circle and a parabola, use the general tangent equation for the parabola and apply the tangency condition with the circle.

Answer 1

The Correct Option is B

Step 1: Equation of the Given Circle 
The given equation of the circle is: \[ x^2 + y^2 = 9. \] This represents a circle centered at \( (0,0) \) with radius \( 3 \). 
Step 2: Equation of the Given Parabola 
The given equation of the parabola is: \[ y^2 = 8x. \] This is a standard parabola of the form \( y^2 = 4ax \), where \( 4a = 8 \), so \( a = 2 \). The focus of this parabola is at \( (2,0) \). 
Step 3: Finding the Common Tangent 
The equation of a common tangent to the circle and the parabola is derived using the standard tangent equation approach. A general tangent to the parabola is given by: \[ y = m x + \frac{2}{m}. \] To also satisfy the tangency condition with the circle, we equate the perpendicular distance from the center \( (0,0) \) to this line with the radius \( 3 \): \[ \frac{|c|}{\sqrt{1 + m^2}} = 3. \] Solving for \( m \) and \( c \), we obtain the required common tangent: \[ x - \sqrt{3}y + 6 = 0. \] 
Final Answer: \( \boxed{x - \sqrt{3}y + 6 = 0} \)