By graphing the given equations, we can determine the number of points of intersection between the circle and the line. To do so, we need to graph both figures. Let's consider one of them at a time.
Graphing the Circle
Let's start by recalling the standard equation of a circle.
(x- h)^2+(y- k)^2= r^2
Here, the center is the point ( h, k) and the radius is r. We will rewrite the given equation to match this form, and then we can identify the center and the radius.
(x-2)^2+(y+3)^2=18
⇕
(x- 2)^2+(y-( - 3))^2=( sqrt(18))^2
The center of the circle is the point ( 2, - 3), and its radius is sqrt(18). We can graph the circle using this information.
Graphing the Line
To graph the line, we will need its equation to be in slope-intercept form to help us identify the slope m and y-intercept b. Fortunately, our equation is already in slope-intercept form. Let's rewrite it a bit, so that the slope and y-intercept are more visible.
y=-2x-2 ⇔ y= -2x+( -2)
We can see that the slope is - 2 and the y-intercept is (0, - 2). To graph this equation, we will start by plotting its y-intercept. Then, we will use the slope to determine another point that satisfies the equation, and connect the points with a line.
Finding the Points of Intersection
We can see that the circle and the line intersect at exactly two points. Let's identify them.
From, the graph we can see that one point of intersection is (-1,0) and the other one seems to occur at (2.6,- 7.2). In order to check whether the second point really is the solution we can solve the system of equations algebraically.
(x-2)^2+(y+3)^2=18 & (I) y=- 2x-2 & (II)
Since the variable y is isolated in Equation (II), let's use Substitution Method to solve the above system.
Now to solve Equation (I) for x, we will use the Quadratic Formula.
x=- b±sqrt(b^2-4ac)/2a
In our case a= 5, b= - 8, and c= - 13. We can substitute these values into the formula and simplify.
The solutions for this equation are x= 8± 1810. Let's separate them into the positive and negative cases.
x=8± 18/10
x_1=8+18/10
x_2=8-18/10
x_1=26/10
x_2=- 10/10
x_1=2.6
x_2=- 1
Using the Quadratic Formula, we found that the solutions of the given equation are x_1=2.6 and x_2=- 1. Finally using Equation (II), we can find the corresponding y-values.
x
y=- 2x-2
Simplify
x= 2.6
y=- 2( 2.6)-2
y=- 7.2
x= - 1
y=- 2( - 1)-2
y=0
We have confirmed that the points (2.6,- 7.2) and (- 1,0) are the points of intersection between the circle and the line.
