Problem 44 - Precalculus

Question

Determine the center and the radius for the circle. Also, find the \(y\) -coordinates of the points (if any) where the circle intersects the \(y\) -axis. $$x^{2}+y^{2}-10 x+2 y+17=0$$

Step-by-Step Solution

Answer

The circle's center is (5, -1) with a radius of 3. It does not intersect the y-axis.

Step 1: Rearrange the Equation

Write the given equation of the circle in the standard form. Begin by grouping the terms involving the same variable. The equation given is \(x^{2}+y^{2}-10x+2y+17=0\). Separate the \(x\) and \(y\) terms: \((x^2 - 10x) + (y^2 + 2y) = -17\).

Step 2: Complete the Square for x

Complete the square for the \(x\) terms. For \(x^2 - 10x\), find the number to complete the square: \(\left(\frac{-10}{2}\right)^2 = 25\). Add and subtract 25: \((x^2 - 10x + 25 - 25)\). This gives \((x-5)^2 - 25\).

Step 3: Complete the Square for y

Complete the square for the \(y\) terms. For \(y^2 + 2y\), find the number to complete the square: \(\left(\frac{2}{2}\right)^2 = 1\). Add and subtract 1: \((y^2 + 2y + 1 - 1)\). This gives \((y+1)^2 - 1\).

Step 4: Rewrite the Equation

Substitute the completed squares back into the equation: \((x-5)^2 - 25 + (y+1)^2 - 1 = -17\). Simplify the equation: \((x-5)^2 + (y+1)^2 = 9\).

Step 5: Identify the Center and Radius

The equation \((x-5)^2 + (y+1)^2 = 9\) is in the form \((x-h)^2 + (y-k)^2 = r^2\), where \((h, k)\) is the center and \(r\) is the radius. Here, the center is \((5, -1)\) and the radius \(r\) is 3 (since \(r^2 = 9\), \(r = \sqrt{9} = 3\)).

Step 6: Find Intersection with y-axis

To find the intersection with the y-axis, set \(x = 0\) in the circle's equation: \((0-5)^2 + (y+1)^2 = 9\). Simplify to \(25 + (y+1)^2 = 9\). Then \((y+1)^2 = -16\), which has no real solutions since a square cannot be negative. Therefore, there are no points where the circle intersects the y-axis.

Key Concepts

Understanding the Center of a Circle

The center of a circle is a crucial concept when dealing with circle equations and geometry. In a standard circle equation form \((x - h)^2 + (y - k)^2 = r^2\), the center is represented by the point \((h, k)\). This point defines the exact middle of the circle in a Cartesian coordinate system.

To find the center, you need to rewrite the circle equation into the standard form. For the example equation \(x^2 + y^2 - 10x + 2y + 17 = 0\), we rearrange it by completing the square for both \(x\) and \(y\) terms.

• First, rearrange the equation: \((x^2 - 10x) + (y^2 + 2y) = -17\).
• Next, complete the square for \(x\): \((x-5)^2 - 25\).
• Then, complete the square for \(y\): \((y+1)^2 - 1\).

After substitution and simplification, we arrive at the equation \((x-5)^2 + (y+1)^2 = 9\). Here, the center \(h, k\) is \(5, -1\).

Calculating the Radius of a Circle

The radius is the distance from the center of the circle to any point on the circle. In our standard circle equation \((x - h)^2 + (y - k)^2 = r^2\), the radius \(r\) is simply the square root of the number on the right side of the equation.

For the example \((x-5)^2 + (y+1)^2 = 9\), you can see that \(r^2 = 9\). Therefore, the radius is:

• \(r = \sqrt{9} = 3\)

The radius helps in understanding the size of the circle and plays a pivotal role in its properties, like circumference and area.

Understanding the Intersection with the y-axis

To determine if and where a circle intersects the y-axis, you need to set \(x = 0\) in the circle's equation because the y-axis corresponds to all points where \(x\) equals zero.

For the equation \((x-5)^2 + (y+1)^2 = 9\), substitute \(x = 0\) to get:

• \((0-5)^2 + (y+1)^2 = 9\)
• Simplify to: \(25 + (y+1)^2 = 9\)
• Thus, \((y+1)^2 = -16\)

A negative result, as found here \((y+1)^2 = -16\), indicates that there are no real solutions to this equation. Therefore, the circle does not intersect the y-axis, as a square of a real number cannot be negative. This insight is essential for understanding the spatial relationship between geometric shapes and the axes they may interact with.