Circle Equation: Tangent To Negative Axes

Oct 10, 2025 by ADMIN 42 views

Alright, guys, let's dive into a cool problem involving circles, tangent lines, and a little bit of algebra. We're tasked with finding the equation of a circle, given some specific conditions. This isn't just about plugging numbers into a formula; it's about understanding the geometry and how the different elements relate to each other. So, buckle up, and let's get started!

Understanding the Problem

First, let's break down the problem statement. We need to find the equation of a circle. Remember, the general equation of a circle is (x - h)^2 + (y - k)^2 = r^2, where (h, k) is the center of the circle, and r is the radius. We're given two key pieces of information:

The center of the circle lies on the line 2x - 4y = 0.
The circle is tangent to both the negative x-axis and the negative y-axis.

Let's visualize this. Imagine the coordinate plane. The negative x-axis is the left side of the x-axis, and the negative y-axis is the bottom part of the y-axis. For a circle to be tangent to both of these axes, its center must be in the third quadrant (where both x and y coordinates are negative). Also, because it's tangent to both axes, the distance from the center to each axis must be the same. This distance is, of course, the radius of the circle. This means that the absolute values of the x and y coordinates of the center are equal, i.e., |h| = |k| = r. Since the center is in the third quadrant, both h and k are negative, so we can say h = -r and k = -r.

Now, this is where the first piece of information comes into play. The center (h, k) lies on the line 2x - 4y = 0. This means that the coordinates of the center must satisfy this equation. We can substitute h and k into the equation to find a relationship between them. Replacing x with h and y with k, we get 2h - 4k = 0.

Solving for the Center and Radius

So far, we have two important equations:

h = -r
k = -r
2h - 4k = 0

We can use these equations to solve for h, k, and r. Let's substitute h = -r and k = -r into the third equation:

2(-r) - 4(-r) = 0

-2r + 4r = 0

2r = 0

r = 0

Wait a minute! r = 0 doesn't make sense. A circle with a radius of 0 is just a point, not a circle tangent to the axes. This indicates that there's most likely an error in the problem setup or in our interpretation. The most probable issue is the equation of the line where the center lies. The equation 2x - 4y = 0 simplifies to x = 2y. This suggests that for any value of y, the x-coordinate is twice as large. This line passes through the first and third quadrants. However, the problem specifies that the circle is tangent to the negative x and y axes, implying that the center must be in the third quadrant, where both coordinates are negative. Let's re-examine the equation of the line and consider a more general approach.

Instead of assuming h = -r and k = -r directly, let's keep h and k as variables and use the tangency condition to relate them to the radius. If the circle is tangent to the negative x-axis, the y-coordinate of the center (k) must be equal to -r (since k is negative in the third quadrant). Similarly, if the circle is tangent to the negative y-axis, the x-coordinate of the center (h) must also be equal to -r (since h is negative). So, we still have h = -r and k = -r.

Let’s backtrack and re-evaluate the given equation of the line: 2x - 4y = 0. This simplifies to x = 2y. Thus, h = 2k.

Now we have:

h = 2k
h = -r
k = -r

From equations (2) and (3), we see that h = k, since both are equal to -r. However, this contradicts equation (1), which states that h = 2k. The only way for both h = k and h = 2k to be true simultaneously is if h = 0 and k = 0. But if h = 0 and k = 0, then r = 0, which, as we discussed, leads to a degenerate circle (a point).

This indicates a fundamental inconsistency in the problem statement. It's impossible for a circle to satisfy all the given conditions simultaneously. The line 2x - 4y = 0 (or x = 2y) simply doesn't allow for a circle tangent to both negative axes to have its center on that line. The equation of the line must be different. We need to consider that there may have been a typo in the original equation.

Let's assume that the correct equation is 2x + 4y = 0 (or x + 2y = 0). In this case:

h + 2k = 0
h = -r
k = -r

Substituting h = -r and k = -r into the first equation:

-r + 2(-r) = 0

-r - 2r = 0

-3r = 0

r = 0

Again, we end up with r = 0, suggesting that either the center of the circle cannot lie on a line that passes through the origin (0,0), or there is indeed an error in the equation of the line itself.

Let us consider a line that does not pass through the origin. Let us assume that the line is $2x + 4y + c = 0$ . Now the equations are:

$2h + 4k + c = 0$
$h = -r$
$k = -r$

Substituting the values of $h$ and $k$ in equation 1:

$2(-r) + 4(-r) + c = 0$

$-2r - 4r + c = 0$

$-6r + c = 0$

$c = 6r$

So, the equation of the line is $2x + 4y + 6r = 0$ . Therefore, the equation of the circle is $(x + r)^2 + (y + r)^2 = r^2$ .

Expanding this gives:

$x^2 + 2rx + r^2 + y^2 + 2ry + r^2 = r^2$

$x^2 + y^2 + 2rx + 2ry + r^2 = 0$

This is the equation of the circle under the assumption that the line is $2x + 4y + 6r = 0$ .

Conclusion

In summary, the original problem presents an inherent inconsistency. A circle tangent to both the negative x-axis and negative y-axis cannot have its center on the line 2x - 4y = 0. We explored various scenarios, including adjustments to the line equation, but each led to a degenerate case (r = 0). This highlights the importance of carefully examining problem statements for feasibility and consistency. We showed that if we assume the line to be $2x + 4y + 6r = 0$ , then the equation of the circle can be determined. Remember, in math, sometimes the most valuable lesson is recognizing when a problem is not well-defined! Keep practicing, guys, and don't be afraid to question the given information.