Problem 68 - Algebra and Trigonometry with Analytic Geometry

Question

Exer. 67-68: For the given circle, find (a) the \(x\)-intercepts and (b) the \(y\)-intercepts. $$ x^{2}+y^{2}-10 x+4 y+13=0 $$

Step-by-Step Solution

Answer

The \(x\)-intercepts are \(5 \pm 2\sqrt{3}\); there are no \(y\)-intercepts.

Step 1: Rewrite Circle Equation in Standard Form

Start with the circle equation \(x^{2}+y^{2}-10x+4y+13=0\). To find intercepts, we first complete the square for both \(x\) and \(y\) terms to rewrite it in standard form \((x-h)^2 + (y-k)^2 = r^2\).

Step 2: Complete the Square for x

Focus on the \(x\) terms: \(x^2 - 10x\). Complete the square by adding and subtracting \(25\): \(x^2 - 10x + 25 - 25\). This becomes \((x-5)^2 - 25\).

Step 3: Complete the Square for y

Focus on the \(y\) terms: \(y^2 + 4y\). Complete the square by adding and subtracting \(4\): \(y^2 + 4y + 4 - 4\). This becomes \((y+2)^2 - 4\).

Step 4: Rewrite the Equation

Substitute the completed squares back into the equation: \((x-5)^2 - 25 + (y+2)^2 - 4 + 13 = 0\). Simplify to get: \((x-5)^2 + (y+2)^2 = 16\).

Step 5: Find x-intercepts

For \(x\)-intercepts, set \(y = 0\) in the standard form. Substitute \(y = 0\) into \((x-5)^2 + (y+2)^2 = 16\) which gives \((x-5)^2 + 4 = 16\). Simplify to \((x-5)^2 = 12\). Solve for \(x\) to find \(x = 5 \pm 2\sqrt{3}\).

Step 6: Find y-intercepts

For \(y\)-intercepts, set \(x = 0\) in the standard form. Substitute \(x = 0\) into \((x-5)^2 + (y+2)^2 = 16\) which gives \(25 + (y+2)^2 = 16\). Simplify to \((y+2)^2 = -9\). Since this cannot yield real solutions, there are no \(y\)-intercepts.

Key Concepts

x-intercepts

X-intercepts are the points where the graph of an equation crosses the x-axis. To find them for a circle, set the y-coordinate to zero because the points lie directly on the x-axis. This reduces the circle equation to a simpler form that only has the variable x.

In this example, the circle equation \[(x-5)^2 + (y+2)^2 = 16\] is used. By setting y = 0, the equation becomes: \[(x-5)^2 + 4 = 16\].

Simplifying this gives \[(x-5)^2 = 12\]. Taking the square root of both sides, we find two potential x-intercepts: \(x = 5 + 2\sqrt{3}\) and \(x = 5 - 2\sqrt{3}\).

Thus, the x-intercepts are the points - \((5 + 2\sqrt{3}, 0)\)
- \((5 - 2\sqrt{3}, 0)\).

Understanding x-intercepts involves setting y = 0 and solving for x, revealing where the circle intersects the x-axis.

y-intercepts

Y-intercepts occur where the graph crosses the y-axis, meaning the x-coordinate is zero at these points. For a circle's equation, setting x to zero will simplify the equation and help determine the y-intercepts.

In the provided circle equation \[(x-5)^2 + (y+2)^2 = 16\], setting x = 0 yields \[25 + (y+2)^2 = 16\]. Simplifying this leads to \[(y+2)^2 = -9\].

Since the square of a number can't be negative and thus cannot equal -9, there are no real y-intercepts. This means that the circle doesn't cross the y-axis.

Y-intercepts are determined by setting x = 0 and examining the resulting equation. Here, the absence of real solutions indicates no points of intersection on the y-axis. This can sometimes happen with circles, especially when they sit to one side of the y-axis.

completing the square

Completing the square is a mathematical technique used to convert quadratic expressions or equations into a perfect square form. This process makes it easier to work with the equation, especially for graphing or solving purposes, such as finding the center and radius of a circle.

To complete the square for the equation \(x^2 - 10x\), focus solely on the x terms. Add and subtract \(25\) (which is \((10/2)^2\)), turning the expression into \((x - 5)^2 - 25\).

For the y terms \(y^2 + 4y\), adding and subtracting \(4\) (since \((4/2)^2\)), results in \((y + 2)^2 - 4\).

Substitute these completed squares back into the original equation to transform it to the standard form: \[(x-5)^2 + (y+2)^2 = 16\].

This crucial step is often used to bring a circle's equation from its general form to the standard form, simplifying the identification of the center point and radius.

standard form of a circle equation

The standard form of a circle equation is a streamlined version that makes it easy to identify the circle's center and radius. The standard format is \((x-h)^2 + (y-k)^2 = r^2\), where \((h, k)\) represents the center of the circle, and \(r\) is the radius.

In this problem, after completing the square, we converted the given circle equation into standard form: \[(x-5)^2 + (y+2)^2 = 16\].

From this, it's clear that the center of the circle is at \((5, -2)\), and the radius is \(\sqrt{16} = 4\).

Using the standard form improves comprehension and allows for quick determination of the circle's geometric properties without additional calculation. Understanding this form is vital for not just circle-related calculations but also graphing them effectively.