Problem 69 - College Algebra

Question

Graph both equations in the same rectangular coordinate system and find all points of intersection. Then show that these ordered pairs satisfy the equations. $$ \begin{array}{r} (x-2)^{2}+(y+3)^{2}=4 \\ y=x-3 \end{array} $$

Step-by-Step Solution

Answer

The points of intersection of the line and circle are (-1, -4) and (2, -1).

Step 1: Graph the first equation

The first equation represents a circle with center at (2,-3) and radius 2. To plot this, start at the point (2, -3) and draw the circle of radius 2.

Step 2: Graph the second equation

The second equation represents a line with slope 1 and y-intercept -3. Start from the y intercept (which is -3) and then for each increase of 1 in x, increase y by 1.

Step 3: Find points of intersection

The points of intersection between the line and the circle are the points that are both on the circle and line. This can be achieved by solving the two equations simultaneously. Hence, substitute \(y\) from the equation of the line (\(y = x - 3\)) into the equation of the circle giving \((x-2)^{2}+((x - 3) + 3 )^{2} = 4\), that simplifies to \(x^{2} - x - 2 = 0\). Solving this quadratic for \(x\) yields the values \(x = -1, 2\), and substituting these into the line equation gives the \(y\) values corresponding to \(x\) which are \(y = -4, -1\). So, the points of intersection are (-1, -4) and (2, -1)

Step 4: Verify that intersection points fit both equations

Substitute each point into both the circle equation and the line equation. For example, taking point (-1, -4), we find that both equations are fulfilled. Repeat this for the other point (2, -1), and confirm it satisfies both equation as well. This verifies that the points are indeed intersections of both geometric shapes.

Key Concepts

Graphing Circles

Graphing a circle involves understanding its equation, which is usually written in the form \((x - h)^2 + (y - k)^2 = r^2\). Here,

• \((h, k)\) is the center of the circle,
• \(r\) is the radius.

For the equation \((x-2)^2+(y+3)^2=4\), the circle is centered at \((2,-3)\) with a radius of \(2\) (since \(\sqrt{4} = 2\)). To graph this,
find the center point, then draw a circle with a radius extending 2 units from that center in all directions.
This visualizes how the circle is plotted in a coordinate system.Understanding this makes it easier to analyze how other geometric figures might intersect with it.

Graphing Lines

Graphing a line involves using the slope-intercept form \(y = mx + b\), where

• \(m\) represents the slope of the line,
• \(b\) is the y-intercept where the line crosses the y-axis.

For the equation \(y = x - 3\), the slope \(m = 1\) tells us that for every increase of 1 in \(x\), \(y\) increases by 1. The y-intercept \(b = -3\)
means the line crosses the y-axis at \((0, -3)\).
Starting at the point \((0, -3)\), you can plot the line by following the slope upward, forming a straight line through the graph. This straightforward process allows us to predict where the line might meet other shapes.

Intersection Points

Finding the intersection points of different geometric figures involves solving their equations simultaneously. In this context:

• The circle's equation is \((x-2)^2+(y+3)^2=4\).
• The line's equation is \(y = x - 3\).

By substituting \(y = x - 3\) into the circle's equation, it becomes a single variable equation.
Solving \((x-2)^2+((x-3)+3)^2=4\) simplifies to \(x^2 - x - 2 = 0\).
Solve this quadratic to find \(x = -1\) and \(x = 2\). Substituting these x-values back to find \(y\), we derive the intersection points \((-1, -4)\) and \((2, -1)\).
This approach helps identify where two entities cross each other on a graph.

Solving Quadratic Equations

Solving quadratic equations is a key skill in determining intersection points for systems of equations. A quadratic equation generally has the form \(ax^2 + bx + c = 0\).Several methods exist:

• Factoring the equation, if possible, into simpler terms.
• Using the quadratic formula: \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\), which calculates the roots of the equation.
• Completing the square, transforming the equation into a perfect square trinomial.

In our example, solving \(x^2 - x - 2 = 0\) (derived from graphing problems), we can factor it as \((x - 2)(x + 1) = 0\) to find \(x = 2\) or \(-1\). Understanding how to solve these equations is essential for identifying where lines and curves intersect on a graph.