What is the simple way of solving quadratic equation
1 answer:
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Answer:
The graph of the line and the circle is in the attachment.
Step-by-step explanation:
In order to graph the circle, you have to use the circle equation formula to obtain the center and the radius.
(x-xo)² + (y-yo)² = r²
where the point (xo,yo) is the center and r is the radius.
Using the given formula, the center is:
xo=1, yo=2
(1,2)
And the radius is:
r=2
Now you have to graph the point (1,2) and draw a circle centered at that point, with a distance of 2 from the center.
To obtain the exact values of the intersection of the circle with the x-axis you have to replace y=0 in the equation and solve it for x. For the points of intersection with the y-axis you have to replace x=0 in the equation and solve it for y.
-For x=0, solving for y:
(0-1)²+(y-2)²=2²
1+(y-2)²=4
(y-2)²= 3
Applying the squareroot both sides
y-2 = ± √3
y1= -√3 +2
y2= √3 +2
P1(0, -√3 +2) and P2(0, √3 +2)
-For y=0
(x-1)²+(0-2)²=2²
(x-1)² + 4 = 4
(x-1)² = 0
Applying the squareroot both sides:
x-1 = √0
x-1=0
x=1
P3 (1,0)
In order to graph the linear equation, you have to obtain the intersection with the x-axis replacing y=0 in the equation and the intersection with the y-axis replacing x=0 in the equation. That way, you obtain two point of the line and then you have to trace the line containing those points.
-For x=0
y=2(0)+2 =2
P1 (0,2)
-For y=0
0= 2x+2
Adding -2 both sides
2x= -2
Dividing by 2 both sides
x= -1
P2(-1,0)
