Consider the two curves C₁ : y² = 4x, C₂ : x² + y² - 6x + 1 = 0. Then, (A) C₁ and C₂ touch each othe

Q: Wait, why does a repeated root mean tangency? Shouldn't it just be a single point?

A repeated root in a quadratic equation $f(x) = 0$ means the curve is tangent to the axis or, in this case, the curves are tangent to each other. It means the intersection points have merged into one, which is the definition of tangency.

Q: Can't we use the distance formula instead of substitution?

You could, but it's overkill. Substitution is the most direct path. JEE is about efficiency, and solving the system of equations is the standard, most robust way to handle intersections.

Q: Why are there two points of tangency?

Because the parabola is symmetric about the x-axis. If it touches the circle at $(1, 2)$, it must also touch it at $(1, -2)$ due to symmetry. The algebra confirms this by giving us $y^2 = 4$, which leads to both positive and negative roots.

Question

Consider the two curves $C_1 : y^2 = 4x, C_2 : x^2 + y^2 - 6x + 1 = 0$. Then,

Options:

(A) $C_1$ and $C_2$ touch each other only at one point

(B) $C_1$ and $C_2$ touch each other exactly at two points

(C) $C_1$ and $C_2$ intersect (but do not touch) at exactly two points

(D) $C_1$ and $C_2$ neither intersect nor touch each other

Sigma Percentile · Accepted Answer

### Visualizing the Curves

We start with the parabola $C_1 : y^2 = 4x$, which is a standard parabola opening to the right with its vertex at the origin.

Then we have the circle $C_2 : x^2 + y^2 - 6x + 1 = 0$. If you complete the square for the circle, you get:

$$(x-3)^2 + y^2 = 8$$

This tells us the circle is centered at $(3, 0)$ with a radius of $\sqrt{8}$. The parabola starts at the origin and opens right, while the circle is centered further along the $x$-axis.

### The Algebraic Bridge

To find the intersection points, we must solve the equations simultaneously. This is our logic bridge.

Since the parabola gives us a direct value for $y^2$, the smartest move is to use the substitution method. We will take $y^2 = 4x$ from the parabola and plug it straight into the circle's equation.

This transforms our two-variable problem into a single-variable quadratic equation.

### The Quadratic Revelation

Substituting $y^2 = 4x$ into the circle's equation $x^2 + y^2 - 6x + 1 = 0$ gives us:

$$x^2 + 4x - 6x + 1 = 0$$

Simplifying this, we get $x^2 - 2x + 1 = 0$. Do you recognize this pattern? It is a perfect square!

We can rewrite this as:

$$(x-1)^2 = 0$$

Solving for $x$, we find $x = 1$. Note that this is a repeated root. In the context of coordinate geometry, a repeated root signifies that the curves just graze or touch each other without crossing. This is the hallmark of **tangency**.

### Geometric Interpretation

Because we found a repeated root at $x = 1$, we know that the parabola and the circle are tangent at this $x$-coordinate.

To find the corresponding $y$-coordinates, we substitute $x = 1$ back into the parabola's equation:

$$y^2 = 4(1) \Rightarrow y^2 = 4$$

Taking the square root, we get $y = 2$ and $y = -2$.

Thus, the curves share exactly two points: **$(1, 2)$ and $(1, -2)$**. At both these points, the curves touch each other. We have successfully navigated the problem, moving from visualization to algebraic manipulation and finally to geometric interpretation.

Animated Solution for Mathematics - Conic Sections: Consider the two curves $C_{1} : y^{2} = 4 x, C_{2} : x^{2} + y^{2} - 6 x + 1 = 0$ . Then,

Select Answer:

Visualized Solution

Visualizing the Curves $C_{1}$ and $C_{2}$

Strategy: Substitution Method

Substituting $y^{2} = 4 x$ into $C_{2}$

Simplifying the Equation

Solving for $x$ : The Repeated Root

Interpreting the Repeated Root

Finding the $y$ -coordinates

Final Conclusion

The Sigma Insight: Standard Equations of Parabola, Ellipse, and Hyperbola

Visualizing the Curves

The Algebraic Bridge

The Quadratic Revelation

Geometric Interpretation

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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; } Consider the two curves C1​:y2=4x,C2​:x2+y2−6x+1=0. Then,