Question

In Problems 65-70, graph each equation and find the point(s) of intersection, if any. The line \(x+2 y=0\) and the circle \((x-1)^{2}+(y-1)^{2}=5\)

Short answer

The points of intersection are \((2, -1)\) and \((-1.2, 0.6)\).

Step-by-step solution

- Rewrite the Line Equation

Start by expressing the line equation in slope-intercept form, which is easier to graph. Rewrite the equation \[ x + 2y = 0 \] Solve for \( y \): \[ y = -\frac{1}{2}x \]

- Identify the Circle's Properties

Examine the equation of the circle: \[ (x-1)^2 + (y-1)^2 = 5 \] It is in standard form \[ (x-h)^2 + (y-k)^2 = r^2 \] where \( (h, k) \) is the center and \( r \) is the radius. Here, the center is at \((1, 1)\) and the radius is \( \sqrt{5} \)

- Substitute Line Equation into Circle Equation

To find the points of intersection, substitute the line equation \( y = -\frac{1}{2}x \) into the circle equation: \[ (x-1)^2 + ((-\frac{1}{2}x) -1)^2 = 5 \] Simplify the equation to find \( x \). \[ (x-1)^2 + (-\frac{1}{2}x - 1)^2 = 5 \] \[ (x-1)^2 + (\frac{1}{4}x^2 + x + 1) = 5 \] \[ x^2 - 2x + 1 + \frac{1}{4}x^2 + x + 1 = 5 \] Combine like terms: \[ \frac{5}{4}x^2 - x + 2 = 5 \]

- Solve the Quadratic Equation

Solve for \( x \) in the quadratic equation \[ \frac{5}{4}x^2 - x + 2 = 5 \] Simplify and solve: \[ \frac{5}{4}x^2 - x - 3 = 0 \] Multiply by 4 to clear the fraction: \[ 5x^2 - 4x - 12 = 0 \] Use the quadratic formula \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \) where \( a = 5 \), \( b = -4 \) and \( c = -12 \): \[ x = \frac{4 \pm \sqrt{16 + 240}}{10} \] \[ x = \frac{4 \pm 16}{10} \] \[ x = 2 \] and \[ x = -1.2 \]

- Find Corresponding y-values

For each \( x \) value found, substitute back into the line equation \( y = -\frac{1}{2}x \): When \( x = 2 \): \[ y = -\frac{1}{2}(2) = -1 \] When \( x = -1.2 \): \[ y = -\frac{1}{2}(-1.2) = 0.6 \]

- Identify the Intersection Points

The points of intersection are at the coordinates: \( (2, -1) \) and \( (-1.2, 0.6) \)

Key concepts

graphing equations

Graphing equations is a key skill in algebra and trigonometry. It helps visualize the relationship between variables. For example, the line equation given in the problem is:
-x + 2y = 0.
To graph this line, you can rewrite it in slope-intercept form as y = -1/2x. This makes it easier to plot. Begin by identifying key points. For instance, an x-value of 2 gives y = -1. Plotting several points and drawing a line through them provides the graph.

Similarly, graphing the circle requires paying attention to its standard equation form, (x-h)^2 + (y-k)^2 = r^2. Here, (h, k) is the center, and r is the radius. For the circle (x-1)^2 + (y-1)^2 = 5, the center is (1, 1), and the radius is √5. Plot the center and measure out the radius in all directions to draw the circle.

points of intersection

Finding points of intersection involves identifying where two graphs meet. For the exercise, we seek where the line and circle intersect.

To find these points, substitute the y-value from the line equation into the circle equation. This substitution results in a quadratic equation that can be solved for x. Once x-values are found, substitute them back into the line equation to determine corresponding y-values. In this problem, solving the equations gives us the points of intersection: (2, -1) and (-1.2, 0.6). These are the coordinates where both the line and circle intersect.

quadratic formula

The quadratic formula is essential for solving quadratic equations like the one we obtained after substituting the line equation into the circle equation. The standard form of a quadratic equation is ax^2 + bx + c = 0. The quadratic formula, x = (-b ± √(b²-4ac))/(2a), gives the solutions directly.

In this exercise, the quadratic equation was 5/4x^2 - x - 3 = 0. By multiplying by 4 to clear the fraction, we had 5x^2 - 4x - 12 = 0. Using the quadratic formula with a = 5, b = -4, and c = -12, we solved for x, yielding the solutions x = 2 and x = -1.2.

circle equations

Circle equations are typically written in the form (x-h)^2 + (y-k)^2 = r^2 where (h, k) denotes the center, and r is the radius. For our problem, the circle equation is (x-1)^2 + (y-1)^2 = 5. This tells us that the circle's center is at (1, 1), and the radius is √5 (since 5 is the square of the radius).

The center and radius can help graph a circle accurately. Circle equations may require manipulation, like completing the square when not in standard form. Here, identifying these parameters quickly reveals both the circle's size and position on the coordinate plane.

line equations

Line equations can come in various forms: general form (Ax + By = C), slope-intercept form (y = mx + b), and point-slope form (y - y1 = m(x - x1)). In this exercise, the given line equation is x + 2y = 0. By converting it to slope-intercept form, y = -1/2x, we make it easier to graph. 

The slope-intercept form highlights the line's slope, m = -1/2, which indicates its steepness and direction. Starting from this form simplifies finding key points along the line. For example, at x = 2, y = -1 gives us one point on the graph. These methods illustrate that converting and graphing line equations leverages their inherent properties for better visualization.