Problem 67 - Algebra and Trigonometry with Analytic Geometry

Question

Exer. 67-68: For the given circle, find (a) the \(x\)-intercepts and (b) the \(y\)-intercepts. $$ x^{2}+y^{2}-4 x-6 y+4=0 $$

Step-by-Step Solution

Answer

The x-intercept is (2, 0). The y-intercepts are (0, 3+√5) and (0, 3-√5).

Step 1: Rewriting the Equation in Standard Form

We start by rewriting the equation of the circle in standard form. The given equation is: \(x^{2}+y^{2}-4x-6y+4=0\). We will complete the square for the \x\ and \y\ components.

Step 2: Completing the Square for x

For the \x\ terms, factor out and rearrange: \(x^{2}-4x\). To complete the square, add and subtract \(\left(\frac{-4}{2}\right)^2 = 4\) inside the equation: \(x^{2} - 4x + 4 - 4\). This becomes \((x - 2)^{2} - 4\).

Step 3: Completing the Square for y

For the \y\ terms, factor out and rearrange: \(y^{2}-6y\). To complete the square, add and subtract \(\left(\frac{-6}{2}\right)^2 = 9\) inside the equation: \(y^{2} - 6y + 9 - 9\). This becomes \((y - 3)^{2} - 9\).

Step 4: Writing the Completed Square Form

Now substitute back into the equation the squares we completed: \((x - 2)^{2} - 4 + (y - 3)^{2} - 9 + 4 = 0\). Simplifying this gives \((x - 2)^{2} + (y - 3)^{2} = 9\), which is the equation of the circle in standard form with center \( (2, 3) \) and radius \3\.

Step 5: Finding the x-intercepts

To find the \x\-intercepts, set \y=0\ in the standard form equation and solve for \x\: \((x - 2)^{2} + (0 - 3)^{2} = 9\). Simplifying gives \((x - 2)^{2} + 9 = 9\), leading to \((x - 2)^{2} = 0\). Thus \x = 2\. The \x\-intercept is \( (2, 0) \).

Step 6: Finding the y-intercepts

To find the \y\-intercepts, set \x=0\ in the standard form equation and solve for \y\: \((0 - 2)^{2} + (y - 3)^{2} = 9\). Simplifying gives \(4 + (y - 3)^{2} = 9\), leading to \( (y - 3)^{2} = 5\). Solving this gives \y - 3 = \pm\sqrt{5}\. Therefore, \y = 3 \pm\sqrt{5}\. The \y\-intercepts are \( (0, 3+\sqrt{5}) \) and \( (0, 3-\sqrt{5}) \).

Key Concepts

Completing the Square

Completing the square is a powerful technique used to rearrange a quadratic equation into a perfect square trinomial. This process is often essential when dealing with equations of circles, primarily when converting them into the standard form.

In the context of a circle equation like \(x^{2} + y^{2} - 4x - 6y + 4 = 0\), the goal is to transform the quadratic components of both \(x\) and \(y\) into neat squares. For \(x^{2} - 4x\), we add and subtract the square of half the linear term coefficient: \(\left(\frac{-4}{2}\right)^2 = 4\). Hence, it becomes \((x - 2)^2 - 4\). Similarly, the \(y\) terms \(y^{2} - 6y\) transform to \((y - 3)^2 - 9\) by adding and subtracting \(\left(\frac{-6}{2}\right)^2 = 9\).

This step helps in reorganizing terms, making further transformations clearer and calculation cleaner.

x-intercepts

Finding the \(x\)-intercepts of a circle involves determining where the circle crosses the \(x\)-axis. At these points, \(y\) is zero.

To find the \(x\)-intercepts from the standard equation \((x - 2)^2 + (y - 3)^2 = 9\), set \(y = 0\). Substituting this into the equation gives \((x - 2)^2 + 9 = 9\) which simplifies to \((x - 2)^2 = 0\). Solving this, we find that \(x = 2\).

Therefore, the \(x\)-intercept of this circle is at the point \((2, 0)\). This process shows the points of intersection effectively by isolating the \(x\) variable.

y-intercepts

The \(y\)-intercepts of a circle are points where the circle crosses the \(y\)-axis, meaning \(x\) is zero.

Using the circle's standard form \((x - 2)^2 + (y - 3)^2 = 9\), set \(x = 0\) to find the \(y\)-intercepts. Substituting gives \(4 + (y - 3)^2 = 9\), which reduces to \((y - 3)^2 = 5\). Solving this equation, we find \(y = 3 \pm \sqrt{5}\).

This results in two \(y\)-intercepts: \((0, 3 + \sqrt{5})\) and \((0, 3 - \sqrt{5})\). Through these calculations, we examine symmetry in a circle's graph about its center, highlighting its intersection with coordinate axes.

Standard Form of a Circle

The standard form of a circle's equation provides a clear view of its fundamental properties: its center and its radius. Often written as \((x - h)^2 + (y - k)^2 = r^2\), where \((h, k)\) is the center and \(r\) is the radius.

Converting \(x^2 + y^2 - 4x - 6y + 4 = 0\) into the standard form involves completing the square for both \(x\) and \(y\). Resulting in \((x - 2)^2 + (y - 3)^2 = 9\), this shows the circle's center at \((2, 3)\) and the radius is \( \sqrt{9} = 3\).

This form not only reveals the circle's location on the coordinate plane but also its size, making further geometric analysis straightforward.

Radius and Center of a Circle

In the standard circle equation \((x - h)^2 + (y - k)^2 = r^2\), the values \(h\), \(k\), and \(r\) are key to understanding the circle's positioning and scale.

The center \((h, k)\) can be directly extracted as \((2, 3)\) from the equation \((x - 2)^2 + (y - 3)^2 = 9\). The radius \(r\) is derived from the equation \(r^2 = 9\), indicating that \(r = 3\).

Knowing these, you can plot the circle accurately on a graph and analyze its geometrical relationships with other figures. This understanding is foundational in geometrical constructions and solving complex geometrical problems effectively.