Consider a circle (x - α)² + (y - β)² = 50, where α, β 0. If the circle touches the line y + x = 0

Q: Wait, why do we reject the negative root $\alpha = -9$?

That is a fantastic question! In coordinate geometry, constraints are not just suggestions; they define the physical reality of the problem. The problem explicitly states that $\alpha, \beta > 0$. This means our circle's center must reside in the first quadrant. If we chose $\alpha = -9$, our center would be in the second quadrant, violating the given conditions. Always check your constraints at the end!

Q: Can we use the distance from the center to the line formula instead of the normal equation?

Absolutely! You could use the formula $d = \frac{|Ax_0 + By_0 + C|}{\sqrt{A^2 + B^2}}$. Here, the distance from $C(\alpha, \beta)$ to $x + y = 0$ is the radius $\sqrt{50}$. So, $\frac{|\alpha + \beta|}{\sqrt{1^2 + 1^2}} = \sqrt{50}$. This gives $|\alpha + \beta| = 10$. Since $\alpha, \beta > 0$, $\alpha + \beta = 10$. Squaring both sides gives $100$. Both methods are elegant; the normal method helps you visualize the geometry, while the distance formula is a powerful algebraic shortcut.

Q: Why is the normal slope $1$?

Great catch. The tangent line is $y = -x$, which can be written as $y = -1x + 0$. The slope $m_t$ is $-1$. Since the normal is perpendicular to the tangent, its slope $m_n$ must satisfy $m_t \cdot m_n = -1$. Therefore, $(-1) \cdot m_n = -1$, which simplifies to $m_n = 1$. It is a fundamental property of circles that the normal at the point of tangency always passes through the center.

Question

Consider a circle $(x - \alpha)^2 + (y - \beta)^2 = 50$, where $\alpha, \beta > 0$. If the circle touches the line $y + x = 0$ at the point $P$, whose distance from the origin is $4\sqrt{2}$, then $(\alpha + \beta)^2$ is equal to

Sigma Percentile · Accepted Answer

### The Geometry of Perfection: Unraveling the Circle

Welcome, my dear student. Today, we are not just solving a problem; we are embarking on a journey through the elegant world of coordinate geometry.

Imagine a circle, a perfect, symmetrical entity, resting in the first quadrant of the Cartesian plane. It is defined by the equation:

$$ (x - \alpha)^2 + (y - \beta)^2 = 50 $$

We know its center is at $C(\alpha, \beta)$ and its radius squared is $50$. We are given a tangent line, $y + x = 0$, which acts as a boundary. Let us uncover the secrets of this circle together.

### Phase 1: The Point of Contact

First, let us focus on the point $P$, where the circle touches the line $y = -x$. We are told that the distance from the origin $O(0,0)$ to this point $P$ is $4\sqrt{2}$.

Since $P$ lies on the line $y = -x$, its coordinates must be of the form $(x, -x)$. Using the distance formula:

$$ \sqrt{x^2 + (-x)^2} = 4\sqrt{2} $$

Squaring both sides, we get $2x^2 = (4\sqrt{2})^2 = 32$. This simplifies beautifully to $x^2 = 16$, giving us $x = \pm 4$.

We have two candidates for $P$: $(4, -4)$ and $(-4, 4)$. Given that our center $C(\alpha, \beta)$ is in the first quadrant, the geometry dictates that we should work with $P(-4, 4)$.

### Phase 2: The Normal Line

Now, here is where the magic happens. In the realm of circles, the normal line—the line perpendicular to the tangent at the point of contact—is the path of truth. It always passes through the center of the circle.

The slope of our tangent line $y = -x$ is $m_t = -1$. Therefore, the slope of the normal line must be the negative reciprocal, $m_n = 1$.

Using the point-slope form with our point $P(-4, 4)$, the equation of the normal becomes:

$$ y - 4 = 1(x - (-4)) \Rightarrow y = x + 8 $$

This is the line upon which our center $C(\alpha, \beta)$ must reside.

### Phase 3: The Algebraic Lock

Since the center $C(\alpha, \beta)$ lies on the line $y = x + 8$, we have the crucial relationship $\beta = \alpha + 8$. Now, we return to the circle's definition.

The distance from the center $C$ to the point of tangency $P$ is the radius, $r = \sqrt{50}$. Thus, $CP^2 = 50$. Substituting our coordinates:

$$ (\alpha - (-4))^2 + (\beta - 4)^2 = 50 $$

Substituting $\beta = \alpha + 8$ into this, we find:

$$ (\alpha + 4)^2 + ((\alpha + 8) - 4)^2 = 50 $$

This simplifies to $(\alpha + 4)^2 + (\alpha + 4)^2 = 50$, or $2(\alpha + 4)^2 = 50$. Dividing by $2$, we get $(\alpha + 4)^2 = 25$.

Taking the square root, $\alpha + 4 = \pm 5$. This gives us $\alpha = 1$ or $\alpha = -9$. Since $\alpha > 0$, we must choose $\alpha = 1$. Consequently, $\beta = 1 + 8 = 9$.

### The Final Celebration

We have arrived at the destination. We have found the center $C(1, 9)$.

The problem asks for the value of $(\alpha + \beta)^2$. Substituting our values:

$$ (\alpha + \beta)^2 = (1 + 9)^2 = 10^2 = \mathbf{100} $$

It is a beautiful, clean result. Remember, geometry is not just about numbers; it is about understanding the relationships between lines, curves, and points. You have mastered this today.

Animated Solution for Mathematics - Circles: Consider a circle $(x - α)^{2} + (y - β)^{2} = 50$ , where $α, β > 0$ . If the circle touches the line $y + x = 0$ at the point $P$ , whose distance from the origin is $42$ , then $(α + β)^{2}$ is equal to

Enter Numerical Value:

Visualized Solution

Analyze the Given Information

Locate Point $P$

Distance Formula for $O P$

Solve for Coordinates of $P$

The Concept of Normal

Equation of Normal at $P (- 4, 4)$

Center on the Normal

Applying the Radius Condition

Distance Formula for $C P$

Simplify the Equation

Solve for $α$

Apply Constraints for Center

Final Calculation

The Sigma Insight: Equation of Tangent and Normal

The Geometry of Perfection

Phase 1

Phase 2

Phase 3

The Final Celebration

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