Question

Determine which of the given points are on the graph of the equation. $$ \begin{array}{l} \text { Equation: } x^{2}+y^{2}=4 \\ \text { Points: }(0,2) ;(-2,2) ;(\sqrt{2}, \sqrt{2}) \end{array} $$

Short answer

The points (0, 2) and (\sqrt{2}, \sqrt{2}) are on the graph. (-2, 2) is not.

Step-by-step solution

Understand the Equation

The given equation is: \[ x^{2} + y^{2} = 4 \] This is the equation of a circle centered at the origin (0, 0) with a radius of 2.

Verify Point (0, 2)

Substitute the coordinates (0, 2) into the equation: \[ 0^{2} + 2^{2} = 4 \] Simplify it to verify: \[ 0 + 4 = 4 \] Since 4 = 4, the point (0, 2) lies on the graph of the equation.

Verify Point (-2, 2)

Substitute the coordinates (-2, 2) into the equation: \[ (-2)^{2} + 2^{2} = 4 \] Simplify it to verify: \[ 4 + 4 = 8 \] Since 8 ≠ 4, the point (-2, 2) does not lie on the graph of the equation.

Verify Point (\sqrt{2}, \sqrt{2})

Substitute the coordinates (\sqrt{2}, \sqrt{2}) into the equation: \[ (\sqrt{2})^{2} + (\sqrt{2})^{2} = 4 \] Simplify it to verify: \[ 2 + 2 = 4 \] Since 4 = 4, the point (\sqrt{2}, \sqrt{2}) lies on the graph of the equation.

Key concepts

Coordinate Geometry

In coordinate geometry, we deal with geometric figures like points, lines, and curves using an algebraic approach. Each point in the plane is described by an ordered pair of numbers \((x, y)\). Equations, like \(\(x^{2} + y^{2} = 4\)\), help us visualize shapes on a coordinate plane. Understanding the relationship between algebraic equations and geometric figures is crucial. The equation given represents a circle, and we need to analyze and understand its properties using coordinate geometry.

Graphing

Graphing equations on a coordinate plane helps visualize and understand them better. For example, let's graph the circle defined by the equation \(\(x^{2} + y^{2} = 4\)\).

**Steps for Graphing a Circle:**

• Identify the center of the circle. In this case, it's (0, 0).
• Determine the radius. Since \(4 = r^{2}\), \(r = 2\).
• Plot points at a distance of 2 units from the center in all directions.
• Draw a smooth, continuous curve through these points.

This visual representation helps in verifying if a given point lies on the circle.

Point Verification

To check if a point lies on the graph of an equation, substitute the point's coordinates into the equation and see if it satisfies the equation.

**Example:**

• For the point (0, 2), substitute \(x = 0\) and \(y = 2\) into the equation. If both sides equal, the point is on the circle.
• For the point (-2, 2), substitute \(x = -2\) and \(y = 2\). If the resulting equation is not true, the point is not on the circle.
• Repeat the process for other points to verify if they lie on the graph.

This method ensures accurate and systematic verification of points.

Circle

A circle is a set of all points in a plane that are equidistant from a fixed point known as the center. The distance from the center to any point on the circle is called the radius.

**Key Properties:**

• The general form of a circle with center \((h, k)\) and radius \(r\) is \((x - h)^{2} + (y - k)^{2} = r^{2}\).
• For our specific problem, the circle's center is \((0, 0)\) and the radius is \(2\).
• This specific circle equation simplifies to \(x^{2} + y^{2} = 4\).

Understanding these properties helps in analyzing and graphing circles effectively.

Radius

The radius is an essential part of understanding circles.

**Key Points:**

• In our equation \(x^{2} + y^{2} = 4\), the radius \(r\) can be derived as \(2\) because \(4 = r^{2}\), so \(r = \sqrt{4} = 2\).
• The radius is the constant distance from the center to any point on the circle.
• Knowing the radius helps in graphing the circle and verifying the points on it.

Remember, the radius is always positive and is a crucial factor in defining the circle.

Substitution Method

The substitution method involves replacing variables with their respective values to verify solutions. Here is how to use this method for point verification:

**Example:**

• Take the point \((0, 2)\) and substitute into the circle's equation: \(0^{2} + 2^{2} = 4\). This simplifies to \(0 + 4 = 4\), confirming the point lies on the circle.
• Try the point \((-2, 2)\): \((-2)^{2} + 2^{2} = 4\). This simplifies to \(4 + 4 = 8\), which does not equal \(4\), so the point is not on the circle.
• For the point \((\sqrt{2}, \sqrt{2})\): \((\sqrt{2})^{2} + (\sqrt{2})^{2} = 4\). This simplifies to \(2 + 2 = 4\), confirming the point lies on the circle.

This step-by-step method helps confirm the accuracy of solutions efficiently.