Circle Equation: Center (-1,1) Through (-3,-1)

Welcome, math enthusiasts! Today, we're diving into the fascinating world of circles and how to find the equation of a circle when we're given its center and a point it passes through. This is a fundamental skill in analytic geometry, and understanding it will unlock many other concepts. Our specific challenge is to determine the equation of a circle that has its center precisely at the coordinates $(-1,1)$ and conveniently passes through the point $(-3,-1)$. This seemingly simple problem involves a few key steps that beautifully illustrate the relationship between a circle's center, its radius, and the coordinates of any point lying on its circumference. We'll break down the process, making it easy to follow and apply to similar problems. Get ready to explore how algebraic equations can perfectly describe these elegant geometric shapes!

Understanding the Standard Equation of a Circle

The journey to finding our circle's equation begins with a solid understanding of its standard form. The standard equation of a circle is a powerful algebraic representation that encapsulates all the information about a circle's position and size. It's derived directly from the Pythagorean theorem and the definition of a circle itself: the set of all points equidistant from a central point. If a circle has its center at coordinates $(h, k)$ and a radius of length $r$, then any point $(x, y)$ on the circle must satisfy the following equation: $(x - h)^2 + (y - k)^2 = r^2$. Here, $(h, k)$ represents the coordinates of the center, and $r$ is the radius. The term $(x - h)^2$ represents the horizontal distance squared from the center to a point on the circle, while $(y - k)^2$ represents the vertical distance squared. Their sum, according to the Pythagorean theorem, is equal to the square of the distance between the center and the point, which is precisely the radius squared. This equation is your key to unlocking circle problems, as it allows you to determine unknown properties like the radius or center if others are known, or to identify the circle itself given its defining characteristics. We will be using this standard form as our foundation to solve the problem at hand, meticulously plugging in the known values and calculating the missing piece: the radius.

Calculating the Radius: The Missing Piece

Our mission is to find the equation of the circle, and we already have half the puzzle: the center $(h, k) = (-1, 1)$. The other crucial piece of information we need for the standard equation is the radius, $r$. Fortunately, the problem provides us with a point that the circle passes through, which is $(-3, -1)$. The distance between the center of the circle and any point on its circumference is, by definition, the radius. Therefore, we can calculate the radius by finding the distance between our center $(-1, 1)$ and the given point $(-3, -1)$. We use the distance formula, which is itself derived from the Pythagorean theorem: $d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$. In our case, $(x_1, y_1)$ will be our center $(-1, 1)$ and $(x_2, y_2)$ will be the point on the circle $(-3, -1)$. Plugging these values into the distance formula, we get: $r = \sqrt{(-3 - (-1))^2 + (-1 - 1)^2}$. Let's simplify this expression. First, calculate the differences in the x and y coordinates: $(-3 - (-1)) = -3 + 1 = -2$, and $(-1 - 1) = -2$. Now, square these differences: $(-2)^2 = 4$ and $(-2)^2 = 4$. Summing these squares gives us $4 + 4 = 8$. Finally, take the square root to find the radius: $r = \sqrt{8}$. For clarity and ease of use in the circle's equation, we can simplify $\sqrt{8}$ to $2\sqrt{2}$. However, for the standard equation of a circle, we actually need $r^2$, the radius squared. So, $r^2 = (\sqrt{8})^2 = 8$. This value, 8, is the square of our radius, and it's exactly what we need to complete our circle's equation. The calculation of the radius is a critical step, as it defines the size of our circle and is essential for its unique algebraic representation.

Assembling the Final Equation

Now that we have successfully determined both the center $(h, k)$ and the square of the radius $r^2$, we can confidently assemble the final equation of the circle. Our center is $(h, k) = (-1, 1)$, and we found that $r^2 = 8$. We substitute these values directly into the standard equation of a circle: $(x - h)^2 + (y - k)^2 = r^2$. Replacing $h$ with $-1$, $k$ with $1$, and $r^2$ with $8$, we get: $(x - (-1))^2 + (y - 1)^2 = 8$. Simplifying the term $(x - (-1))$, we change it to $(x + 1)$. Therefore, the final, simplified standard equation of the circle is $(x + 1)^2 + (y - 1)^2 = 8$. This equation is a complete and precise description of the circle. Any point $(x, y)$ that satisfies this equation lies on the circumference of the circle with center $(-1, 1)$ and a radius whose square is 8. This process elegantly demonstrates how basic geometric principles and algebraic manipulation come together to define and describe geometric shapes. We've successfully navigated from given information to a definitive equation, showcasing the power of analytical geometry.

Verifying Our Solution

To ensure our hard work has paid off and our equation is indeed correct, it's always a good practice to perform a verification of the solution. We found the equation of the circle to be $(x + 1)^2 + (y - 1)^2 = 8$. We know that the circle must pass through the point $(-3, -1)$. Let's substitute the coordinates of this point into our derived equation and see if it holds true. So, we set $x = -3$ and $y = -1$. The equation becomes: $(-3 + 1)^2 + (-1 - 1)^2$. Let's evaluate this expression. First, the terms inside the parentheses: $(-3 + 1) = -2$ and $(-1 - 1) = -2$. Now, we square these results: $(-2)^2 = 4$ and $(-2)^2 = 4$. Adding these squared values gives us $4 + 4 = 8$. Since the left side of our equation evaluates to 8, and the right side of our equation is also 8, the equation holds true: $8 = 8$. This confirms that the point $(-3, -1)$ indeed lies on the circle described by the equation $(x + 1)^2 + (y - 1)^2 = 8$. This verification step is crucial for building confidence in our results and ensuring accuracy in our mathematical problem-solving. It reinforces the understanding that the equation perfectly represents the geometric object it describes.

Conclusion: Mastering Circle Equations

In conclusion, we have successfully determined the equation of a circle with center $(-1,1)$ that passes through the point $(-3,-1)$. By leveraging the standard equation of a circle, $(x - h)^2 + (y - k)^2 = r^2$, and applying the distance formula to calculate the radius, we arrived at the precise equation: $(x + 1)^2 + (y - 1)^2 = 8$. This process highlights the interplay between algebra and geometry, where algebraic formulas provide a powerful lens through which to understand and describe geometric shapes. Mastering these foundational concepts in analytic geometry is vital for further exploration in mathematics, from understanding conic sections to advanced calculus. The ability to translate geometric properties into algebraic equations, and vice versa, is a hallmark of mathematical proficiency. Keep practicing these types of problems, and you'll find yourself becoming increasingly comfortable and adept at solving them. Remember, understanding the core principles is key to unlocking more complex mathematical challenges. Keep exploring the beautiful world of mathematics!

For further exploration on circles and conic sections, you can visit mathworld.wolfram.com.