Find The Circle Equation: Center (0,-9), Point (-√56,-7)

When dealing with circles in geometry, understanding how to determine their equation of the circle is fundamental. A circle is defined by its center and its radius. If we know these two pieces of information, we can easily write down the standard equation of the circle. The standard form of the equation of a circle with center $(h, k)$ and radius $r$ is given by $(x-h)^2 + (y-k)^2 = r^2$. In this article, we'll walk through how to find the equation of a circle when you're given the coordinates of its center and the coordinates of a point that lies on the circle. This is a common problem in coordinate geometry, and it requires us to use the distance formula, which is derived from the Pythagorean theorem, to find the radius. Let's dive into the specifics of our problem: we have a circle with its center at the point $(0, -9)$ and we know that the point $(-\sqrt{56}, -7)$ is located on the circumference of this circle.

Understanding the Core Concepts: Center, Radius, and the Distance Formula

To find the equation of the circle, we first need to identify the values of $h$, $k$, and $r$. We are already given the center of the circle, which directly provides us with the values for $h$ and $k$. In this case, the center $(h, k)$ is $(0, -9)$, so $h = 0$ and $k = -9$. The trickier part is finding the radius, $r$. Remember, the radius is the distance from the center of the circle to any point on its circumference. Since we are given a point $(-\sqrt{56}, -7)$ that lies on the circle, the distance between this point and the center $(0, -9)$ will be our radius. This is where the distance formula comes into play. The distance $d$ between two points $(x_1, y_1)$ and $(x_2, y_2)$ in a Cartesian coordinate system is given by $d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$. We will use this formula to calculate the distance between our center $(0, -9)$ and the given point $(-\sqrt{56}, -7)$. Once we have this distance, which is our radius $r$, we can substitute $h$, $k$, and $r$ into the standard equation of a circle.

Step-by-Step Calculation to Determine the Radius

Let's meticulously calculate the radius using the distance formula. Our two points are the center $(x_1, y_1) = (0, -9)$ and the point on the circle $(x_2, y_2) = (-\sqrt{56}, -7)$. Plugging these values into the distance formula, we get:

$r = \sqrt{(-\sqrt{56} - 0)^2 + (-7 - (-9))^2}$

First, let's simplify the terms inside the square root:

$(-\sqrt{56} - 0)^2 = (-\sqrt{56})^2 = 56$

And for the y-coordinates:

$(-7 - (-9))^2 = (-7 + 9)^2 = (2)^2 = 4$

Now, substitute these simplified values back into the distance formula:

$r = \sqrt{56 + 4}$

$r = \sqrt{60}$

So, the radius of our circle is $\sqrt{60}$. For the equation of the circle, we actually need $r^2$. Squaring both sides of $r = \sqrt{60}$, we get $r^2 = 60$. This value of $r^2$ is crucial for completing our equation. It represents the square of the distance from the center to any point on the circle. The calculation involved ensuring we correctly handled the negative signs and the square root. The intermediate steps are important to avoid errors, especially when dealing with square roots and coordinates that might be negative. This systematic approach ensures accuracy in finding the radius, which is a key component of the circle's equation.

Assembling the Final Equation of the Circle

Now that we have all the necessary components, we can construct the equation of the circle. We know the center $(h, k) = (0, -9)$, which means $h=0$ and $k=-9$. We have also calculated that $r^2 = 60$. Substituting these values into the standard equation of a circle, $(x-h)^2 + (y-k)^2 = r^2$, we get:

$(x - 0)^2 + (y - (-9))^2 = 60$

Let's simplify this equation:

$x^2 + (y + 9)^2 = 60$

This is the standard form of the equation for the circle with center $(0, -9)$ that passes through the point $(-\sqrt{56}, -7)$. The $x^2$ term comes from $(x-0)^2$, which is simply $x^2$. The $(y+9)^2$ term arises from substituting $k=-9$ into $(y-k)^2$, resulting in $(y - (-9))^2$, which simplifies to $(y+9)^2$. The right side of the equation is $r^2$, which we found to be 60. This equation precisely defines all the points $(x, y)$ that are equidistant from the center $(0, -9)$, with that distance being $\sqrt{60}$. The process involved correctly identifying the center coordinates, applying the distance formula to find the radius squared, and then substituting these values into the standard circle equation format. Each step builds upon the previous one, ensuring that the final equation accurately represents the described circle. It's a direct application of fundamental geometric principles in an algebraic context.

Verification and Interpretation

To ensure our equation of the circle is correct, we can perform a quick verification. We know the point $(-\sqrt{56}, -7)$ should satisfy the equation $x^2 + (y + 9)^2 = 60$. Let's substitute $x = -\sqrt{56}$ and $y = -7$ into the equation:

$(-\sqrt{56})^2 + (-7 + 9)^2$

$= 56 + (2)^2$

$= 56 + 4$

$= 60$

Since the left side equals the right side (60), our equation is indeed correct. The equation $x^2 + (y + 9)^2 = 60$ represents a circle centered at the origin of the x-axis and shifted down by 9 units along the y-axis. The radius squared is 60, meaning the radius is $\sqrt{60}$, which is approximately 7.75. This means that any point on the circle is exactly $\sqrt{60}$ units away from the center $(0, -9)$. The point $(-\sqrt{56}, -7)$ is one such point. This exercise highlights the power of coordinate geometry in describing geometric shapes using algebraic equations. Understanding the relationship between the center, radius, and points on the circle is key to solving such problems. It also demonstrates how the distance formula, derived from the Pythagorean theorem, is a versatile tool in coordinate geometry for calculating distances between points.

For further exploration into circles and their equations, you can visit resources like MathWorld's Circle Page or Khan Academy's Geometry Section.