Problem 13 - Algebra and Trigonometry Real Mathematics, Real People

Question

Identify the center and radius of the circle. $$x^{2}+y^{2}=36$$

Step-by-Step Solution

Answer

The center of the circle is at the point (0, 0) and its radius is 6 units.

Step 1: Identify the center

The center of the circle is given by the values of \(h\) and \(k\) in the standard form of the circle's equation. In the equation \(x^{2}+y^{2}=36\), no specific values are subtracted from \(x\) or \(y\), which means \(h = 0\) and \(k = 0\). Therefore, the center of the circle is at the origin, \((0, 0)\).

Step 2: Identify the radius

The radius of the circle is given by the value of \(r\) in the standard form of the circle's equation, and is found by taking the square root of the number on the right side of the equation. In the equation \(x^{2}+y^{2}=36\), \(r^{2} = 36\). So, to identify the radius, take the square root of 36, which yields \(r = 6\). Therefore, the radius of the circle is 6 units.

Key Concepts

Standard Form of a Circle

Understanding the standard form of a circle's equation is crucial to solving many problems in algebraic geometry. The standard form is expressed as \( (x - h)^2 + (y - k)^2 = r^2 \), where \( (h, k) \) represents the center of the circle, and \( r \) is the radius. It's akin to a treasure map, where \( h \) and \( k \) tell us the exact location of the buried treasure—the center, and \( r \) tells us how far we need to dig around the center to find all the treasure—representing the circle's boundary.

The equation \( x^2 + y^2 = 36 \) from the exercise may not appear in standard form at first glance; however, since there are no values subtracted from \( x \) and \( y \) it's as though \( h \) and \( k \) are both zero. Hence, this equation is actually in standard form with the center at the origin. To compare, if the equation contained terms like \( -2x \) or \( +3y \) within the parentheses, we would know the values subtracted from \( x \) and \( y \) directly reveal the coordinates of the center.

Radius of a Circle

Think of the radius of a circle as the lifeline connecting the center to the circle's perimeter. It is constant at any point along the boundary and is what most circular attributes derive from. In algebraic terms, the radius is often represented by the symbol \( r \). We find it in the standard form equation by equating it to the square root of the constant on the other side of the equation, expressing it as \( r = \sqrt{r^2} \).

When we look at the given exercise, \( x^2 + y^2 = 36 \), we see that \( 36 \) is the value of \( r^2 \). Thus, we calculate \( r \) as \( \sqrt{36} \) which simplifies down to \( r = 6 \). This radius, or 'lifeline', measures 6 units. It's essential to note that the radius can also inform us about other properties of the circle, such as its area ( \( A = \pi r^2 \) ) and circumference ( \( C = 2\pi r \) ).

Center of a Circle

In the context of a circle, the center is the anchor point from which every line segment extending to the circle's edge is exactly the radius length away. In simpler terms, it is the 'middle' of the circle. Algebraically, the center's coordinates are denoted by \( (h, k) \) in the standard form equation.

Taking the exercise \( x^2 + y^2 = 36 \) as an example, since there are no numbers subtracting from \( x \) and \( y \) in the equation, we quickly ascertain the center is at \( (0, 0) \), the origin of the coordinate plane. This central spot is our starting point for all measurements of the circle. If there were numbers subtracting from \( x \) and \( y \) in the equation—say \( (x - 3)^2 \) or \( (y + 4)^2 \)— our center would be at \( (3, -4) \), respectively.

Algebraic Geometry

Algebraic geometry is the branch of mathematics that marries algebra, specifically polynomial equations, with geometric concepts like points, lines, and curves. This field allows us to describe and analyze geometric shapes using algebraic expressions, effectively bridging the gap between numeric and visual understanding.

For instance, circles, which we are focused on with our exercise \( x^2 + y^2 = 36 \), are geometric shapes that can be described in the Cartesian coordinate system using algebraic equations. These equations provide concrete formulas for us to deduce properties such as the circle's size and location on a plane. By converting geometric questions into algebraic problems, algebraic geometry provides us with a powerful toolkit for tackling a wide array of problems in both pure and applied mathematics.