Question

Stellt \(x(6-3 x)+3 y(12-y)=72\) einen Kreis dar? Welchen Mittelpunkt und Radius hat er gegebenenfalls.

Short answer

The given equation does represent a circle. The circle's center is at point (1,2) and it has a radius of \( \sqrt{29} \) units.

Step-by-step solution

Structure Equation

The goal is to structure the equation \( x(6-3x)+3y(12-y) = 72 \) into the standard form of a circular equation \( (x-h)^2 + (y-k)^2 = r^2 \). For that, group the x-terms and y-terms and sort them out: \( x^2(3)-(6)x + y^2(3)-(12)y = 72 \). Divide everything by 3 and you get: \( x^2 - 2x + y^2 - 4y = 24 \)

Complete the Square

Add and subtract the square of half the coefficient of x and the square of half the coefficient of y to the left side to format the equation into a circle's standard form: \( (x-1)^2 - 1 + (y-2)^2 - 4 = 24 \). Simplify to get: \( (x-1)^2 + (y-2)^2 = 29 \)

Identify Radius and Centre

From the equation found in step 2, you can identify the center (h, k) and the radius (r). Here, h=1, k=2, and \( r=\sqrt{29} \)

Key concepts

Complete the Square

Understanding how to 'complete the square' is essential for working with quadratic equations and, importantly, for dealing with circle equations in algebra. To 'complete the square' means to take a quadratic expression of the form ax^2 + bx + c and convert it into a perfect square trinomial, (x + d)^2, which makes it easier to solve or graph the equation.

Here's how you do it in a nutshell: first, you ensure that the coefficient of the squared term is 1. If not, you'd divide the entire equation by this coefficient. Then, you take half of the coefficient of the x-term, square it, and add and subtract this value from the equation. This new element fundamentally transforms the equation, capturing the essence of the relationship in a tidy package called a perfect square trinomial.

In our exercise, after dividing by 3, we have x^2 - 2x + y^2 - 4y = 24. To complete the square for the x-terms, we take half of -2, which is -1, square it to get 1, and add and subtract it inside the equation. We do the same for the y-terms, taking half of -4 to get -2, square it to get 4, and add and subtract it inside the equation. This results in (x-1)^2 - 1 + (y-2)^2 - 4 = 24, perfectly setting us up to identify the circle's characteristics.

Identify Circle Center

To identify the center of a circle from its equation, you should first make sure that the equation is in the standard circle form: (x-h)^2 + (y-k)^2 = r^2. Here, (h, k) represents the coordinates of the circle's center.

The process of completing the square allows us to rewrite the given equation in such a way that these values, h and k, become obvious. In our example, after completing the square, we have (x-1)^2 + (y-2)^2 = 29. This tells us directly that the circle is centered at (1, 2) because we see that from the equation, h = 1 and k = 2.

Identifying the circle's center is a crucial step because it gives us a reference point to graph the circle and understand its position relative to the coordinate plane.

Radius of a Circle

The radius of a circle in an equation formatted as (x-h)^2 + (y-k)^2 = r^2 is given by the square root of the right side of the equation, r. The radius reveals the size of the circle and is the distance from any point on the circle to its center.

Continuing from the example above, we simplified the equation to (x-1)^2 + (y-2)^2 = 29 after completing the square. To find the radius, we take the square root of 29, yielding r = √29. Remember, the radius should be expressed as a positive value because it represents a distance. The radius is instrumental in sketching the circle and understanding its scope within the coordinate space.