Problem 23 - Algebra 2

Question

For each equation, find the center and radius of the circle. $$ x^{2}+(y+3)^{2}=25 $$

Step-by-Step Solution

Answer

The center of the circle is at (0, -3), and the radius of the circle is 5.

Step 1: Identify the coordinates of the center

In the general formula for a circle's equation, \((h, k)\) represents the center. Here, we see that x doesn't have any value subtracted or added to it, and \(y + 3\) is squared. Hence, the center of the circle is at the point (0,-3). The sign of the y-coordinate is negative because of the negative sign in the formula.

Step 2: Determine the radius

In the general formula for a circle's equation, \(r^2\) represents the square of the radius. Given that our equation is equal to 25, we know that \(r^2 = 25\). Taking the square root of 25, we find the radius to be 5. It is important to note that radius cannot be negative, so we neglect the negative root.

Key Concepts

Center of a Circle

Understanding the center of a circle is crucial when dealing with circle equations. The general form of a circle's equation is \( (x - h)^2 + (y - k)^2 = r^2 \). The center of the circle is found at the point \((h, k)\), where the variables \(h\) and \(k\) are the shifts applied to the \(x\) and \(y\) coordinates respectively.

If, for example, an equation looks like\(x^2 + (y + 3)^2 = 25\),notice that there is no shift in the \(x\) coordinate, implying that \(h = 0\). The term \((y + 3)^2\) indicates a shift of 3 units downward on the \(y\) axis, so \(k = -3\). Therefore, \((h, k)\) which is \((0, -3)\) is the circle's center point. Remember, the signs in the equation are opposite of the center's values due to the way the equation is structured.

Radius of a Circle

The radius of a circle is a key component of its equation, represented as \(r\)in the general equation:\((x - h)^2 + (y - k)^2 = r^2\). The circle's radius is a measure of the distance from the center to any point on the circle. It is always a positive value.

In the formula, \(r^2\) is easily identified. Consider \(x^2 + (y + 3)^2 = 25\). Here, \(r^2 = 25\). To find the radius \(r\), take the square root of \(25\), which gives a radius \(r = 5\). Remember that radius is always non-negative, so we ignore the negative square root by convention. Being aware of \(r^2\) allows us to effortlessly understand and work with circles.

General Equation of a Circle

The general equation of a circle in the Cartesian plane provides immense insight into its geometry. This equation is typically presented in the form\((x - h)^2 + (y - k)^2 = r^2\),where:

• \(h\) and \(k\) are the \(x\) and \(y\) coordinates of the circle's center, respectively.
• \(r\) is the radius of the circle.

Every equation parameter affects the circle's position and size.

When examining an equation like \(x^2 + (y + 3)^2 = 25\),it does not explicitly present \(h\) and \(k\). The absence of \(x\) associated shifts indicates \(h = 0\). The added \(3\) with \(y\) suggests \(k = -3\). The equation equates to \(25\), hence \(r^2 = 25\). This informs us the circle's radius is \(5\). As every part of the equation is dissected, the circle's true form is revealed, allowing students to visualize and calculate any attributes necessary.