Reflection Of The Circle X²+y²= 4 Across The Line Y = -V

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Hey guys, have you ever wondered what happens when you reflect a circle across a line? Today, we're diving deep into the fascinating world of geometric transformations, specifically focusing on reflecting the circle X²+y²= 4 across the line y = -V. This might sound a bit complex at first, but trust me, we'll break it down step by step so you can master this concept. Let's embark on this mathematical adventure together and uncover the secrets behind this transformation!

Understanding the Basics

Before we jump into the specifics of reflecting our circle, let's make sure we're all on the same page with the fundamental concepts. Think of reflection as creating a mirror image of a shape or object. Imagine placing a mirror on a line; the reflection is what you'd see on the other side of that mirror. In mathematical terms, this line is called the line of reflection. When we reflect a point across a line, the reflected point is the same distance from the line as the original point, but on the opposite side. This line acts as the perpendicular bisector of the segment connecting the original point and its reflection. So, if you fold the paper along the reflection line, the original point and its reflected image would perfectly overlap – pretty neat, right?

Now, let's talk about the equation X²+y²= 4. What does this represent? Well, this is the equation of a circle centered at the origin (0, 0) with a radius of 2. To visualize this, picture a perfectly round circle sitting right in the middle of your coordinate plane. The radius, which is the distance from the center to any point on the circle, is 2 units. Remember that the general equation of a circle centered at (h, k) with radius r is (x - h)² + (y - k)² = r². In our case, h and k are both 0, and r² is 4, which means r is the square root of 4, or 2. Got it? Great! We're building a solid foundation here.

Finally, let's consider the line y = -V. This is a horizontal line that intersects the y-axis at the point -V. Imagine a straight line running parallel to the x-axis, but it's located V units below the x-axis (assuming V is a positive number). The value of V will determine how far down this line sits on the coordinate plane. Now that we've refreshed these basics, we're well-prepared to tackle the reflection of our circle. We know what reflection means, we understand the equation of our circle, and we're clear on what the line of reflection looks like. Let's move on to the exciting part – the transformation itself!

Performing the Reflection

Alright, let's get to the heart of the matter: reflecting the circle X²+y²= 4 across the line y = -V. The key to understanding this transformation lies in how the y-coordinates change while the x-coordinates remain the same. Remember, we're reflecting across a horizontal line, so the horizontal position of each point on the circle won't be affected. It's the vertical position, the y-coordinate, that will change. Think of it like flipping the circle vertically over the line y = -V.

To find the reflected image, we need to determine how the y-coordinate of each point on the circle transforms. Let's say we have a point (x, y) on the original circle. When reflected across the line y = -V, the new y-coordinate, let's call it y', will be such that the line y = -V is the midpoint between y and y'. In other words, the distance from y to -V is the same as the distance from -V to y'. Mathematically, we can express this relationship as:

(-V - y) = (y' - (-V))

Simplifying this equation, we get:

-V - y = y' + V

Now, we can solve for y':

y' = -y - 2V

This is the crucial transformation rule for the y-coordinate. For any point (x, y) on the original circle, its reflected image will have the coordinates (x, -y - 2V). The x-coordinate stays the same, and the new y-coordinate is calculated using this formula. Let's pause for a moment and make sure this makes sense. We're essentially taking the original y-coordinate, flipping it across the x-axis (which is the -y part), and then shifting it down by 2V units (the -2V part). This shift accounts for the distance between the x-axis and our line of reflection, y = -V.

Now that we have the transformation rule, we can apply it to the equation of the circle. The original equation is X²+y²= 4. We need to replace y with our transformed y-coordinate, which is -y - 2V. This will give us the equation of the reflected circle. So, we substitute y' = -y - 2V into the equation. However, remember that the x-coordinate doesn't change during this transformation. Therefore, when finding the equation of the transformed circle, we directly substitute the expression for the new y-coordinate (y') in terms of the old y-coordinate into the original equation. This means we'll manipulate the equation we derived, y' = -y - 2V, to express y in terms of y'. This gives us y = -y' - 2V. Now we substitute this expression for y into the original circle equation:

X² + (-y' - 2V)² = 4

This is the equation of the reflected circle! We've successfully transformed the original circle across the line y = -V. Let's simplify this equation to get a clearer picture of what it looks like.

Determining the Image

Okay, we've arrived at the equation of the reflected circle: X² + (-y' - 2V)² = 4. Now, let's simplify this to better understand the image we've created. First, we can expand the squared term:

X² + (y' + 2V)² = 4

Remember that squaring a negative term results in a positive term, so (-y' - 2V)² is the same as (y' + 2V)². Now, let's expand this binomial:

X² + (y'² + 4Vy' + 4V²) = 4

This is the equation of the reflected circle in terms of x and y'. To make it look cleaner, we can drop the prime notation on y' and simply use y:

X² + (y² + 4Vy + 4V²) = 4

Or, we can write it as:

X² + y² + 4Vy + 4V² = 4

Now, let's analyze this equation. Notice that the X² + y² terms are still present, which indicates that we still have a circle. The presence of the 4Vy term tells us that the center of the reflected circle is no longer at the origin. In fact, the center has shifted vertically. To find the center and radius of the reflected circle, we need to rewrite the equation in the standard form of a circle's equation: (x - h)² + (y - k)² = r², where (h, k) is the center and r is the radius.

To do this, we'll complete the square for the y terms. We have y² + 4Vy. To complete the square, we need to add and subtract (4V/2)² = (2V)² = 4V² inside the equation. Our equation becomes:

X² + (y² + 4Vy + 4V²) = 4

We already have 4V² on the left side, so we can rewrite the equation as:

X² + (y + 2V)² = 4

Now, this equation is in the standard form of a circle's equation! We can clearly see that the center of the reflected circle is (0, -2V) and the radius is √4 = 2. So, what does this tell us? The reflection has shifted the center of the circle from (0, 0) to (0, -2V), but the radius has remained the same. This makes sense, as reflections preserve the size and shape of the object being reflected. The entire circle has simply been moved vertically downwards by a distance of 2V units.

In summary, when the circle X²+y²= 4 is reflected across the line y = -V, the image is another circle with the same radius (2) but with its center shifted to (0, -2V). The equation of the reflected circle is X² + (y + 2V)² = 4. We've successfully determined the image of the circle after the reflection. Awesome job, guys!

Visualizing the Transformation

To truly grasp the concept of this reflection, let's visualize what's happening on the coordinate plane. Imagine our original circle, X²+y²= 4, sitting comfortably centered at the origin (0, 0). It has a radius of 2, so it extends 2 units in all directions from the center. Now, picture the line y = -V running horizontally across the plane, V units below the x-axis. This is our line of reflection – our