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: \( \boxed{x - \sqrt{3}y + 6 = 0} \)

Answer 2

To find the common tangent to the circle \( x^2 + y^2 = 9 \) and the parabola \( y^2 = 8x \), we use the method of comparing slopes and finding tangent equations.

Equation of the Circle:

The circle, \( x^2 + y^2 = 9 \), has its center at \( (0,0) \) and radius \( 3 \).

Equation of the Parabola:

The parabola is \( y^2 = 8x \).

Slope Matching:

The equation of a tangent to the circle at any point can be assumed \( y = mx + c \). Since it is tangent, distance from the center of the circle to the line should be \( r \) (where \( r = 3 \)).

Using the general tangent form for a parabola \( y = mx + \frac{a}{m} \), where \( y^2 = 4ax \) and \( a = 2 \), the equation becomes \( y = mx + \frac{2}{m} \). Comparing this with \( y = mx + c \), we equate \( c = \frac{2}{m} \).

Finding m:

Use the distance formula: the perpendicular distance from the center \( (0, 0) \) to the line \( y = mx + c \) is \( \frac{|c|}{\sqrt{1 + m^2}} = 3 \). Thus,

\(\frac{\left|\frac{2}{m}\right|}{\sqrt{1 + m^2}} = 3 \)

\(\Rightarrow \frac{2}{m\sqrt{1 + m^2}} = 3 \)

\(\Rightarrow \frac{2}{3} = m\sqrt{1 + m^2} \)

\(\Rightarrow 4 = 9m^2 + 9m^4 \)

\(\Rightarrow 9m^4 + 9m^2 - 4 = 0 \)

Solving this quadratic in terms of \( m^2 \), substitute \( z = m^2 \):

\(9z^2 + 9z - 4 = 0 \)

Using the quadratic formula, \( z = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \):

\(z = \frac{-9 \pm \sqrt{81 + 144}}{18} \)

\(z = \frac{-9 \pm 15}{18} \)

This gives \( z = \frac{1}{3} \) (valid since \( z = m^2 \)). Thus, \( m = \pm \frac{1}{\sqrt{3}} \).

Given \( m = \frac{1}{\sqrt{3}} \), then \( c = \frac{2}{m} = 2\sqrt{3} \) or \( m = -\frac{1}{\sqrt{3}}, c = -2\sqrt{3} \).

Verify lines:

Choosing \( y = \frac{1}{\sqrt{3}}x + 2\sqrt{3} \) simplifies to: \( \sqrt{3}y = x + 6 \) which rearranges to \( x - \sqrt{3}y + 6 = 0 \), matching the given correct answer.