Problem 65 - Algebra and Trigonometry

Question

65–72 ? Show that the equation represents a circle, and find the center and radius of the circle. $$ x^{2}+y^{2}-2 x+4 y+1=0 $$

Step-by-Step Solution

Answer

The circle's center is at (1, -2) and its radius is 2.

Step 1: Identify the Standard Circle Equation

The standard form of a circle's equation is given by \((x-h)^2 + (y-k)^2 = r^2\), where \((h, k)\) is the center and \(r\) is the radius. To identify the center and radius, we need to transform the given equation into this standard form.

Step 2: Rearrange the Terms

Start by rearranging the given equation: \[x^2 + y^2 - 2x + 4y + 1 = 0\] to group \(x\) and \(y\) terms:\[x^2 - 2x + y^2 + 4y = -1\].

Step 3: Complete the Square for x

To complete the square for \(x\), take the \(x\) terms:\[x^2 - 2x\]Half the coefficient of \(x\) is \(-1\), and squaring it gives \(1\). Add and subtract \(1\) to get:\[(x^2 - 2x + 1) - 1 = (x-1)^2 - 1\]

Step 4: Complete the Square for y

To complete the square for \(y\), take the \(y\) terms:\[y^2 + 4y\]Half the coefficient of \(y\) is \(2\), and squaring it gives \(4\). Add and subtract \(4\) to get:\[(y^2 + 4y + 4) - 4 = (y+2)^2 - 4\]

Step 5: Express the Equation in Standard Form

Substitute the completed squares back into the equation:\[(x-1)^2 - 1 + (y+2)^2 - 4 = -1\]Simplify the equation:\[(x-1)^2 + (y+2)^2 = 4\]

Step 6: Identify the Center and Radius

Now that the equation is in the standard form, \((x-1)^2 + (y+2)^2 = 4\),we can see the center is \((h, k) = (1, -2)\), and the radius \(r\) is \(\sqrt{4} = 2\).

Key Concepts

Completing the Square

To transform an equation into the standard form of a circle, completing the square is a crucial technique. This method helps to convert quadratic terms into a perfect square trinomial, which can be rewritten as squares of binomials. Let's break this down further:

When you have an equation like \(x^2 - 2x\), the goal is to express it as \((x-h)^2\). Here's how you do it:

• Identify the linear coefficient of \(x\), which is \(-2\) in this example.
• Take half of it, resulting in \(-1\), and then square it to get \(1\).
• Add and subtract this square within the equation to maintain the original balance.

So, \(x^2 - 2x\) becomes \((x-1)^2 - 1\). Now, your \(x\)-terms are neatly wrapped in a square. Apply the same process to the \(y\)-terms for consistent results.

This rearrangement plays a pivotal role in recognizing the center and radius of the circle from its general quadratic form.

Center of a Circle

The center of a circle derived from its equation is the point from which all points on the circle are equidistant. When we rearrange the given circle equation into the standard form \((x-h)^2 + (y-k)^2 = r^2\), the terms \((h, k)\) represent the center's coordinates. Consider our example:

After completing the square, the equation becomes \((x-1)^2 + (y+2)^2 = 4\). From this form:

• \(h = 1\), which is extracted from \((x-h)\).
• \(k = -2\), coming from \((y-k)\).

This tells us the center of the circle is at \((1, -2)\). The steps of completing the square help highlight this crucial information by revealing the circle's symmetry about this central point.

Radius of a Circle

The radius of a circle is the distance from its center to any point on the circle. Once an equation is in the form \((x-h)^2 + (y-k)^2 = r^2\), \(r\) represents the radius. From our transformed equation \((x-1)^2 + (y+2)^2 = 4\), we find the radius by determining \(r^2\):

• The right side of the equation is \(r^2 = 4\).
• Taking the square root solves for \(r\), so \(r = \sqrt{4} = 2\).

Thus, the radius of the circle is \(2\). Understanding and identifying \(r\) in this manner is essential, as it conveys the size of the circle, determining the extent of its reach from the centralized point.