Equation of a common tangent to the circle, x² + y² - 6x = 0 and the parabola, y² = 4x, is: (A) 2√3

Q: Why did we choose the parabola's tangent form instead of the circle's?

Great question! Using the parabola's tangent form $y = mx + \frac{a}{m}$ allows us to parameterize all possible tangents to the parabola using a single variable, the slope $m$. If we started with the circle, we would have to deal with trigonometric parameters like $3 + 3\cos\theta$ and $3\sin\theta$, which often leads to more complex algebraic expressions. Always choose the path that simplifies your variables!

Q: What if the tangent is a vertical line? Does $y = mx + a/m$ cover it?

That is a very sharp observation! The form $y = mx + a/m$ assumes the tangent is not vertical (i.e., $m$ is finite). In this specific problem, a vertical line $x = k$ cannot be tangent to the parabola $y^2 = 4x$ because the parabola extends infinitely to the right. Thus, we are safe to assume the tangent has a finite slope.

Q: Is there a faster way than the distance formula?

You could use the condition that the quadratic equation formed by substituting the line into the circle's equation must have a discriminant of zero (i.e., $D=0$). However, the perpendicular distance formula is generally more robust and less prone to calculation errors when dealing with circle-line tangency. Stick to the distance formula; it is the gold standard for this type of problem.

Question

Equation of a common tangent to the circle, $x^2 + y^2 - 6x = 0$ and the parabola, $y^2 = 4x$, is:

Options:

(A) $2\sqrt{3} y = 12x + 1$

(B) $2\sqrt{3} y = -x - 12$

(C) $\sqrt{3} y = x + 3$

(D) $\sqrt{3} y = 3x + 1$

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine standing on the coordinate plane. To your right, a parabola $y^2 = 4x$ opens its arms wide, and to your left, a circle $x^2 + y^2 - 6x = 0$ sits perfectly centered at $(3, 0)$ with a radius of $3$.

We are looking for a common tangent—a line that gracefully kisses both curves. This is a dance of geometry where we must force a line to satisfy two different constraints simultaneously.

### The Parabola's Secret

We begin by parameterizing our candidate line. For any parabola in the form $y^2 = 4ax$, the equation of a tangent with slope $m$ is given by $y = mx + \frac{a}{m}$.

In our case, comparing $y^2 = 4x$ with the standard form, we find $a = 1$. Thus, any tangent to our parabola must take the form $y = mx + \frac{1}{m}$.

To make this easier to work with, we rewrite it in the general form $Ax + By + C = 0$. Multiplying by $m$, we get $my = m^2x + 1$, or:

$$m^2x - my + 1 = 0$$

This is our 'master key'—a line that is guaranteed to be tangent to the parabola for any non-zero $m$.

### The Circle's Constraint

Now, we turn to the circle. We know its center is $(3, 0)$ and its radius is $3$.

For our candidate line to also be tangent to this circle, the perpendicular distance from the center $(3, 0)$ to the line $m^2x - my + 1 = 0$ must be exactly equal to the radius, $3$.

Using the distance formula $d = \frac{|Ax_1 + By_1 + C|}{\sqrt{A^2 + B^2}}$, we substitute our values:

$$\frac{|m^2(3) - m(0) + 1|}{\sqrt{(m^2)^2 + (-m)^2}} = 3$$

This simplifies to:

$$\frac{|3m^2 + 1|}{\sqrt{m^4 + m^2}} = 3$$

### The Grand Finale

Now, the algebra takes over. We square both sides to eliminate the absolute value and the square root:

$$\frac{(3m^2 + 1)^2}{m^4 + m^2} = 9$$

Cross-multiplying gives us $(3m^2 + 1)^2 = 9(m^4 + m^2)$. Expanding the left side, we get:

$$9m^4 + 6m^2 + 1 = 9m^4 + 9m^2$$

The $9m^4$ terms vanish, leaving us with $6m^2 + 1 = 9m^2$, which simplifies beautifully to $3m^2 = 1$, or $m^2 = \frac{1}{3}$. This gives us two possible slopes: $m = \pm \frac{1}{\sqrt{3}}$.

Choosing the positive slope $m = \frac{1}{\sqrt{3}}$, we substitute it back into our tangent equation:

$$y = \frac{1}{\sqrt{3}}x + \sqrt{3}$$

Multiplying by $\sqrt{3}$, we arrive at the final, elegant equation:

$$\sqrt{3}y = x + 3$$

We have successfully navigated the geometry and found the common tangent. Remember, the key is to assume a tangent for one curve and force it to satisfy the condition for the other. It works every time!

Animated Solution for Mathematics - Conic Sections: Equation of a common tangent to the circle, $x^{2} + y^{2} - 6 x = 0$ and the parabola, $y^{2} = 4 x$ , is:

Select Answer:

Visualized Solution

Visualizing the Curves

Tangent to the Parabola

Substituting Parabola Parameter

Rearranging the Tangent Equation

Analyzing the Circle Geometry

Condition for Tangency to Circle

Applying the Tangency Condition

Simplifying the Distance Equation

Squaring Both Sides

Expanding and Solving for $m^{2}$

Finding the Slope $m$

Substituting $m$ into Tangent Equation

Finalizing the Equation

Conclusion

The Sigma Insight: Equation of Tangent and Normal

Analyzing the Setup

The Parabola's Secret

The Circle's Constraint

The Grand Finale

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Animated Solution for Mathematics - Conic Sections: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Equation of a common tangent to the circle, x2+y2−6x=0 and the parabola, y2=4x, is: