Problem 61 - Precalculus

Question

Find any points of intersection of the graphs of the equations algebraically and then verify using a graphing utility. $$\begin{aligned} &-x^{2}+y^{2}+4 x-6 y+4=0\\\ &x^{2}+y^{2}-4 x-6 y+12=0 \end{aligned}$$

Step-by-Step Solution

Answer

The given equations do not intersect at any point.

Step 1: Rewrite equations

First, rewrite both equations to more standard forms: \[ (x-2)^{2}+(y-3)^{2}=1 \] \[ (x-2)^{2}+(y-3)^{2}=4 \] These are circles centered at (2, 3) with radii of 1 and 2, respectively.

Step 2: Set equations equal to each other

Since we're trying to find the intersection points, we can set the two equations equal to each other: \[ (x-2)^{2}+(y-3)^{2}=1 \] \[ (x-2)^{2}+(y-3)^{2}=4 \] Solving them gives: \[ 0 = 3 \] This is clearly a contradiction, and therefore there are no points of intersection.

Step 3: Verification with graphing utility

After analysing algebraically, use a graphing utility to verify the result. Just plot both of the given equations. It will be evident from the graph that these two circles don't intersect each other, verifying our analysis.

Key Concepts

Graphing Utility Verification

In tackling geometry problems, especially those involving complex shapes like circles, using a graphing utility can be incredibly handy. After solving equations algebraically, graphing utilities serve as a visual confirmation of the solution. By entering the equations of the circles into the utility, you can get an immediate visual representation. If the circles were to intersect, the points of intersection would be where their graphs meet. In our case, the equations represent two circles that, when plotted, would reveal no intersection points. This aligns perfectly with the algebraic finding that no solution exists, demonstrating the power of graphing utilities as a verification tool.

System of Equations

The system of equations for two circles involves finding their points of intersection which are solutions that satisfy both equations simultaneously. When we set the circle equations equal to each other, as shown in Step 2 of the solution, we expect to find common solutions. However, if the resulting solution to this system is a contradiction, as in our exercise resulting in the equation \( 0 = 3 \), we can conclude that the system has no solution, meaning the circles do not intersect at any point. Systems of equations can often be interpreted graphically, where the number of solutions corresponds to the number of intersection points—again highlighting the interconnectedness between algebra and geometry.

Algebraic Method

The algebraic method of solving equations involves manipulating the equations to reveal solutions directly. In the context of intersection points of circles, this method restructures the original equations into a more recognizable form, such as the standard equation of a circle. Step 1 of the solution showcases this by rewriting the given equations into their standard forms, thus making it easier to compare and contrast both equations. Algebraic manipulation can lead to outright solutions, like determining radii and centers, or to contradictions which inform us that there are no intersection points—all without ever having to draw a graph.

Circle Equations

Circle equations are a fundamental aspect of coordinate geometry. They are commonly written in the standard form \( (x-h)^{2} + (y-k)^{2} = r^{2} \), where \( h \) and \( k \) are the x and y coordinates of the circle’s center, and \( r \) is the radius. Understanding the standard form of circle equations is crucial as it immediately gives insights into the properties of the circle - the location of its center and its size. Recognizing that the given equations in the exercise are circles centered at (2, 3), and identifying their radii as 1 and 2 respectively, is paramount to approaching the intersection problem logically and effectively.