Consider a circle C1 : x² + y² - 4x - 2y = α - 5. Let its mirror image in the line y = 2x + 1 be ano

Q: Wait, why do we assume the radius is the same for both circles?

That is a brilliant observation! Reflection is a rigid transformation, also known as an isometry. In geometry, an isometry preserves distances and angles. Since a circle is defined by the set of points at a fixed distance from a center, reflecting the entire set across a line simply moves the center to a new location while keeping the distance from the center to any point on the circumference unchanged. Thus, the radius must be invariant.

Q: Can I just use the distance formula to find the center instead of the reflection formula?

Absolutely! You can find the center $(f, g)$ by noting two things: first, the midpoint of the segment connecting the original center and the new center must lie on the line of reflection. Second, the line connecting the two centers must be perpendicular to the line of reflection. Using these two conditions—midpoint and slope—will lead you to the exact same coordinates as the reflection formula.

Q: Why did we divide the equation of $C_2$ by 5?

Standard form is our best friend in coordinate geometry. The general equation of a circle is $x^2 + y^2 + 2gx + 2fy + c = 0$. If the coefficients of $x^2$ and $y^2$ are not 1, we cannot directly identify the center as $(-g, -f)$ or the radius as $\sqrt{g^2 + f^2 - c}$. Dividing by 5 is the essential first step to normalize the equation so we can extract the geometric properties correctly.

Q: Is it possible for $\alpha$ to be negative?

In this specific problem, $\alpha$ represents the square of the radius ($r^2 = \alpha$). Since the radius of a real circle must be a positive real number, $\alpha$ must be positive. If you ever find yourself with a negative value for $\alpha$ in a similar problem, it usually indicates that the circle is imaginary or that there was a calculation error in the constant term of the circle's equation.

Q: Why is the constant term in the circle equation $\alpha - 5$ and not just $\alpha$?

That is just a way to test your algebraic vigilance! The constant term in the general form $x^2 + y^2 + 2gx + 2fy + c = 0$ is $c$. When you complete the square to find the radius, the constant term moves to the other side of the equation. Always be careful to isolate the constant term correctly before calculating the radius.

Question

Consider a circle $C_1 : x^2 + y^2 - 4x - 2y = \alpha - 5$. Let its mirror image in the line $y = 2x + 1$ be another circle $C_2 : 5x^2 + 5y^2 - 10fx - 10gy + 36 = 0$. Let $r$ be the radius of $C_2$. Then $\alpha + r$ is equal to ________

Sigma Percentile · Accepted Answer

### The Geometry of Reflection: A Journey into Symmetry

Welcome, fellow traveler of the JEE landscape. Today, we are not just solving a coordinate geometry problem; we are exploring the elegant dance of symmetry.

Imagine you are standing in front of a mirror. Your reflection is identical in size, yet flipped in orientation. This is exactly what happens when we reflect a circle across a line.

The circle does not grow, nor does it shrink. It simply shifts its position in the Cartesian plane. This fundamental insight is the key to unlocking our problem.

### Decoding the First Circle

We begin with the equation $x^2 + y^2 - 4x - 2y = \alpha - 5$. To understand this circle, we must bring it into its standard form.

By completing the square for $x$ and $y$, we reveal its soul. We rewrite the equation as:

$$(x - 2)^2 + (y - 1)^2 = \alpha$$

Here, we see the center $O_1$ is at $(2, 1)$ and the radius $r_1$ is $\sqrt{\alpha}$. This is our starting point, our anchor in the coordinate plane.

### The Mirror Transformation

Now, we consider the second circle, $C_2$, defined by $5x^2 + 5y^2 - 10fx - 10gy + 36 = 0$. Before we do anything, we must normalize this.

Dividing by $5$, we get:

$$x^2 + y^2 - 2fx - 2gy + \frac{36}{5} = 0$$

The center $O_2$ is $(f, g)$, and the radius $r$ is $\sqrt{f^2 + g^2 - \frac{36}{5}}$. Because reflection is an isometry, we know with absolute certainty that $r_1 = r$.

This implies the following equality:

$$\sqrt{\alpha} = \sqrt{f^2 + g^2 - \frac{36}{5}}$$

### The Reflection Formula

To find the center $(f, g)$, we use the reflection formula for a point $(x_1, y_1)$ across the line $ax + by + c = 0$. The formula is:

$$\frac{x_2 - x_1}{a} = \frac{y_2 - y_1}{b} = -2 \frac{ax_1 + by_1 + c}{a^2 + b^2}$$

Substituting our center $(2, 1)$ and the line $2x - y + 1 = 0$, we calculate the constant:

$$-2 \frac{2(2) - 1(1) + 1}{2^2 + (-1)^2} = -2 \frac{4}{5} = -\frac{8}{5}$$

Solving for $f$ and $g$ becomes a simple linear exercise:

$$\frac{f - 2}{2} = -\frac{8}{5} \Rightarrow f = -\frac{6}{5}$$

$$\frac{g - 1}{-1} = -\frac{8}{5} \Rightarrow g = \frac{13}{5}$$

We have successfully located the new center $O_2 = (-\frac{6}{5}, \frac{13}{5})$.

### The Grand Finale

With the center known, we return to the radius equation. Substituting our values, we find:

$$r^2 = \left(-\frac{6}{5}\right)^2 + \left(\frac{13}{5}\right)^2 - \frac{36}{5} = \frac{36}{25} + \frac{169}{25} - \frac{180}{25} = \frac{25}{25} = 1$$

Thus, $r = 1$. Since $r_1 = r$, we have $\sqrt{\alpha} = 1$, which means $\alpha = 1$.

The final step is simply the sum:

$$\alpha + r = 1 + 1 = \mathbf{2}$$

Look at the elegance of this result! Through the symmetry of reflection and the rigor of coordinate geometry, we have unraveled the mystery. Keep this clarity with you as you tackle your next challenge.

Enter Numerical Value:

Visualized Solution

Analyze Circle $C_{1}$

Analyze Circle $C_{2}$

The Mirror Image Concept

Reflection Formula Setup

Compute the Reflection Constant

Solve for $f$

Solve for $g$

Calculate Radius $r$ (Part 1)

Calculate Radius $r$ (Part 2)

Find $α$ and Final Answer

The Sigma Insight: Standard and General Equation of a Circle

The Geometry of Reflection

Decoding the First Circle

The Mirror Transformation

The Reflection Formula

The Grand Finale

Similar Questions

Let the mirror image of a circle $c_{1} : x^{2} + y^{2} - 2 x - 6 y + α = 0$ in line $y = x + 1$ be $c_{2} : 5 x^{2} + 5 y^{2} + 10 g x + 10 f y + 38 = 0$ . If $r$ is the radius of circle $c_{2}$ , then $α + 6 r^{2}$ is equal to ________

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Let a circle $C : (x - h)^{2} + (y - k)^{2} = r^{2}, k > 0$ , touch the x-axis at $(1, 0)$ . If the line $x + y = 0$ intersects the circle $C$ at $P$ and $Q$ such that the length of the chord $P Q$ is 2, then the value of $h + k + r$ is equal to ____.

Enter Numerical Value:

Visualized Solution

Analyze Circle .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } C1​

Analyze Circle .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } C2​

The Mirror Image Concept

Reflection Formula Setup

Compute the Reflection Constant

Solve for .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } f

Solve for .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } g

Calculate Radius .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } r (Part 1)

Calculate Radius .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } r (Part 2)

Find .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } α and Final Answer