If the image of the point (-4, 5) in the line x + 2y = 2 lies on the circle (x + 4)² + (y - 3)² = r²

Q: Wait, why do we use the formula $\frac{x - x_1}{a} = \frac{y - y_1}{b} = -2 \frac{ax_1 + by_1 + c}{a^2 + b^2}$? Can't I just find the foot of the perpendicular?

You absolutely can! Finding the foot of the perpendicular is a fundamental method. However, the formula provided is a direct algebraic shortcut derived from that exact process. It saves you time during the exam by combining the slope condition and the midpoint condition into one elegant expression.

Q: Why is there a negative sign in the formula? Does it matter if I write $ax + by + c = 0$ or $ax + by = c$?

The negative sign is crucial because it accounts for the 'reflection' across the line. If you move the constant to the other side, the sign of $c$ changes, which would flip your result. Always ensure your equation is in the form $ax + by + c = 0$ before plugging values into the formula!

Q: I got $r^2 = 4$, so why is the answer 2 and not $\pm 2$?

Great catch on the algebra! In geometry, the radius $r$ represents a physical distance. Just like time or mass in physics, a geometric length must be non-negative. We always take the positive root for the radius.

Q: Is there a way to solve this without finding the exact coordinates of the image point?

While you could potentially use properties of distance from a point to a line, finding the image point is the most robust and least error-prone method here. Sometimes, the 'long' way is actually the 'fast' way because it's more intuitive!

Q: What if the point $P$ was already on the line? What would the image be?

If the point $P$ lies on the line, then $ax_1 + by_1 + c = 0$. If you look at the formula, the entire right-hand side becomes zero, meaning $x = x_1$ and $y = y_1$. The image of a point on the mirror is the point itself! It's a beautiful consistency check.

Question

If the image of the point $(-4, 5)$ in the line $x + 2y = 2$ lies on the circle $(x + 4)^2 + (y - 3)^2 = r^2$, then $r$ is equal to:

Options:

(A) 2

(B) 3

(C) 1

(D) 4

Sigma Percentile · Accepted Answer

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

Imagine you are standing in a dark room, holding a flashlight. You point it at a mirror, and the light bounces off, creating a perfect, symmetrical image on the other side.

In coordinate geometry, we do exactly this with points and lines. Today, we are going to explore the elegant dance of reflection, where a point $P(-4, 5)$ meets the line $L: x + 2y - 2 = 0$ and finds its twin, the image $P'$, which then lands perfectly on a circle.

This isn't just algebra; it's the geometry of symmetry.

### Phase 1: The Mirror Formula

To find the image $P'(x, y)$ of a point $P(x_1, y_1)$ across a line $ax + by + c = 0$, we rely on a powerful geometric truth: the line segment connecting the point and its image must be perpendicular to the mirror line, and the mirror line must bisect that segment.

While we could derive this using slopes and midpoints, we have a beautiful, compact formula at our disposal:

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

This formula is a gift. It encapsulates the entire geometric requirement of reflection into a single ratio.

Let’s prepare our values: $x_1 = -4$, $y_1 = 5$, $a = 1$, $b = 2$, and $c = -2$. Notice how we carefully ensure the line is in the form $ax + by + c = 0$.

If we had written it as $x + 2y = 2$, we might have forgotten to bring the $2$ over, which would have ruined our calculation. Precision is the soul of geometry!

### Phase 2: Calculating the Reflection

Now, let's substitute our values into the formula. First, we calculate the constant ratio on the right side:

$$	ext{Ratio} = -2 \frac{1(-4) + 2(5) - 2}{1^2 + 2^2} = -2 \frac{-4 + 10 - 2}{1 + 4} = -2 \frac{4}{5} = -\frac{8}{5}$$

This ratio is the key that unlocks the coordinates of our image. By equating the $x$ and $y$ components to this ratio, we find the exact location of $P'$.

For $x$, we have $\frac{x + 4}{1} = -\frac{8}{5}$, which gives us $x = -4 - \frac{8}{5} = -\frac{28}{5}$.

For $y$, we have $\frac{y - 5}{2} = -\frac{8}{5}$, which simplifies to $y - 5 = -\frac{16}{5}$, leading us to $y = 5 - \frac{16}{5} = \frac{9}{5}$.

Our image point $P'$ is $(-\frac{28}{5}, \frac{9}{5})$. It feels like we've just tracked a particle through a mirror, doesn't it?

### Phase 3: The Circle's Embrace

Now that we have our point $P'$, we are told it lies on the circle $(x + 4)^2 + (y - 3)^2 = r^2$. This is the final act of our play.

We substitute our coordinates into the circle's equation:

$$\left(-\frac{28}{5} + 4ight)^2 + \left(\frac{9}{5} - 3ight)^2 = r^2$$

Let's simplify the fractions inside the brackets. $-\frac{28}{5} + 4$ becomes $-\frac{28}{5} + \frac{20}{5} = -\frac{8}{5}$.

Similarly, $\frac{9}{5} - 3$ becomes $\frac{9}{5} - \frac{15}{5} = -\frac{6}{5}$. Squaring these, we get:

$$\left(-\frac{8}{5}ight)^2 + \left(-\frac{6}{5}ight)^2 = r^2$$

$$\frac{64}{25} + \frac{36}{25} = r^2$$

Adding these together, we find $\frac{100}{25} = r^2$, which simplifies beautifully to $r^2 = 4$. Since the radius of a circle must be a positive distance, we conclude that **$r = 2$**.

### Conclusion

Look at what we achieved! We started with a point and a line, navigated the reflection, and ended with the radius of a circle.

The beauty of this problem lies in how different branches of mathematics—linear equations, reflections, and circles—converge into a single, elegant result.

Never fear the fractions or the negative signs; they are just the markers on the map leading you to the truth. Keep practicing, keep visualizing, and most importantly, keep enjoying the process of discovery.

Animated Solution for Mathematics - Straight Lines: If the image of the point $(- 4, 5)$ in the line $x + 2 y = 2$ lies on the circle $(x + 4)^{2} + (y - 3)^{2} = r^{2}$ , then $r$ is equal to:

Select Answer:

Visualized Solution

Visualizing the Setup

The Image Formula

Substituting the Values

Evaluating the Constant Ratio

Solving for $x$

Solving for $y$

The Image Point $P^{'}$

Substituting into the Circle

Simplifying the Terms

Calculating $r^{2}$

Finding the Final Radius

Summary and Conclusion

The Sigma Insight: Distance of a Point from a Line

The Mirror of Geometry

Phase 1

Phase 2

Phase 3

Conclusion

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Animated Solution for Mathematics - Straight Lines: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If the image of the point (−4,5) in the line x+2y=2 lies on the circle (x+4)2+(y−3)2=r2, then r is equal to: