Question

Determine the equation of a circle with centre at the origin and radius a) 4 units b) 3 units c) 12 units d) 2.6 units

Short answer

The equations are a) \(x^2 + y^2 = 16\), b) \(x^2 + y^2 = 9\), c) \(x^2 + y^2 = 144\), d) \(x^2 + y^2 = 6.76\)

Step-by-step solution

Understand the Standard Equation of a Circle

The standard equation of a circle with center at the origin \((0,0)\) is given by the formula \(x^2 + y^2 = r^2\), where \(r\) is the radius of the circle.

Determine the Radius for Each Case

Identify the radius given in each part of the problem:a) 4 unitsb) 3 unitsc) 12 unitsd) 2.6 units

Plug in the Radius for Each Part

Substitute the radius into the standard equation for each case:a) For \(r = 4\), the equation is \(x^2 + y^2 = 4^2 = 16\).b) For \(r = 3\), the equation is \(x^2 + y^2 = 3^2 = 9\).c) For \(r = 12\), the equation is \(x^2 + y^2 = 12^2 = 144\).d) For \(r = 2.6\), the equation is \(x^2 + y^2 = 2.6^2 = 6.76\).

Write Down the Final Equations

Each part gives us a specific equation for the circle:a) \(x^2 + y^2 = 16\)b) \(x^2 + y^2 = 9\)c) \(x^2 + y^2 = 144\)d) \(x^2 + y^2 = 6.76\)

Key concepts

Radius

The radius is a crucial part of any circle. It's the distance from the center of the circle to any point on its circumference. For a circle centered at the origin (0,0), the radius is denoted as \(r\). To calculate the radius if we know any point on the circle, use the distance formula. If a point (x, y) lies on the circle, then the radius \( r\) can be found using the equation \[ r = \sqrt{x^2 + y^2} \].
In practical terms, think of the radius as the “reach” of the circle from its central point out to its edge.

Standard Equation

The standard equation of a circle is a mathematical expression that defines all the points (x, y) that make up the circle. For a circle centered at the origin (0,0), the standard equation is given by \[ x^2 + y^2 = r^2 \], where \[ r \] is the radius.
This equation states that for any point on the circle, the sum of the squares of its x and y coordinates will equal the square of the radius. This relationship helps in easily plotting the circle or understanding its geometry.

• It is important because it simplifies many geometric problems involving circles.
• It makes it easy to determine whether a point lies inside, on, or outside the circle.

Center at Origin

In the given problem, all circles are centered at the origin—which is the point (0,0) on the coordinate plane. This makes calculations easier because the formula for the circle’s equation simplifies. Instead of adding or subtracting values to shift the center, the equation remains \[ x^2 + y^2 = r^2 \].
When the center is at the origin, the \( h \) and \( k \) (which represent the x and y coordinates of the circle's center in a more general equation) are both zero. So, the equation \[ (x - h)^2 + (y - k)^2 = r^2 \] simplifies directly into our standard equation.

Circle Equation Examples

To fully understand these concepts, let's look at the specific examples given in the exercise:

• For a radius of 4 units: The equation will be \[ x^2 + y^2 = 4^2 = 16 \].
• For a radius of 3 units: The equation becomes \[ x^2 + y^2 = 3^2 = 9 \].
• With a radius of 12 units: It follows that \[ x^2 + y^2 = 12^2 = 144 \].
• And for a radius of 2.6 units: The equation will be \[ x^2 + y^2 = 2.6^2 = 6.76 \].

These examples illustrate the simplicity and elegance of using the standard equation of a circle.