a To find the number of points of intersection, we will draw the graph of both the circle and the parabola on the same coordinate grid. Let's start with the circle.
Graphing the Circle
To identify the center and the radius of the given circle, we will recall the standard equation of a circle.
(x- h)^2+(y- k)^2= r^2
In the standard equation h is the x-coordinate of the center, k is the y-coordinate of the center, and r is the radius of the circle. Now, let's take a look at the given equation in order to identify the center and the radius.
x^2 + y^2 = 25
⇕
(x- 0)^2 + (y- 0)^2 = 5^2
The center of the given circle is ( 0, 0) and its radius is 5. Let's use this information to draw a graph of the circle.
Graphing the Parabola
To graph the parabola we first need to identify the coefficients of the parabola a, b, and c using the given equation.
y=x^2 - 13
⇕
y= 1x^2+ 0x+( - 13)
For this equation we have that a= - 1, b= - 2, and c= - 2. Now we can find the vertex using its formula. To do this we will need to think of y as a function of x, y=f(x).
Vertex of a Parabola: ( - b/2 a,f(- b/2 a) )
Let's find the x-coordinate of the vertex.
The y-coordinate of the vertex is - 13. Thus, the vertex is at the point (0,- 13). With this we also know that the axis of symmetry of the parabola is the line x=0. Next, let's find two more points on the curve: one on each side of the axis of symmetry.
x
x^2-13
y=x^2-13
-2
( -2)^2-13
- 9
2
2^2-13
- 9
Both ( -2,- 9) and ( 2,- 9) are on the graph. Let's form the parabola by connecting these points and the vertex with a smooth curve.
Let's graph both the circle and the parabola on the same coordinate plane.
Points of Intersection
Looking at the obtained graph, we can observe how many points of intersection of the circle and the parabola there are.
There are exactly 4 points of intersection of the given curves.
b To find the exact coordinates of the points of intersection of the given circle and parabola, we need to write the system of equations composed of the given formulas.
x^2+y^2=25 & (I) y=x^2-13 & (II)
We can solve the system of equations using the Substitution Method. To do so, let's isolate x^2 in Equation (I). Then, we will substitute its value in Equation (II).
Note that in Equation (II) we have a quadratic equation in terms of only the y-variable.
y^2+y-12=0
⇕
1y^2+ 1y+( -12)=0
To solve the quadratic equation, we can substitute a= 1, b= 1, and c= - 12 into the Quadratic Formula.
This result tells us that we have two solutions, y= -1± 72. One of them will use the positive sign and the other one will use the negative sign.
y=-1± 7/2
y=-1+7/2
y=-1-7/2
y=6/2
y=-8/2
y=3
y=-4
Now, consider Equation (I).
x^2 = 25 - y^2
We can substitute both y=3 and y=- 4 into the above equation to find the values for x. Let's start with y=3.
We found that x = ± 4 when y=3. These two solutions of the system, which are points of intersection of the circle and the parabola, are (4,3) and (-4, 3). To find the other solutions, we will substitute -4 for y in Equation (I) again.
We found that x=± 3 when y=- 4. Therefore the next two solutions, which are the other points of intersection of the circle and the parabola, are (3, -4) and (-3, -4).
