Analyzing Arc Equations - Why are x1,x2 & y1,y2 Half Circles?

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SUMMARY

The discussion centers on the mathematical derivation of arc equations from the general circle formula, specifically how the equations for x1, x2, y1, and y2 represent half circles. The equations derived are x1,2 = a ± √(r² - (y - b)²) and y1,2 = b ± √(r² - (x - a)²), which illustrate the left and right arcs for x and the top and bottom arcs for y, respectively. The explanation emphasizes the importance of understanding the coordinate system where x measures horizontal displacement and y measures vertical displacement, clarifying the geometric interpretation of these equations.

PREREQUISITES
  • Understanding of the general circle equation (x-a)² + (y-b)² = r²
  • Basic knowledge of algebraic manipulation and solving equations
  • Familiarity with coordinate geometry concepts
  • Knowledge of square roots and their implications in graphing
NEXT STEPS
  • Study the derivation of the circle equation from basic geometric principles
  • Learn about the properties of arcs and their equations in coordinate geometry
  • Explore the implications of solving quadratic equations in two dimensions
  • Investigate the graphical representation of circles and arcs using graphing software
USEFUL FOR

Students of mathematics, educators teaching geometry, and anyone interested in understanding the relationship between circle equations and their graphical representations.

kishtik
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The first general circle formula is,
[tex] (x-a)^2+(y-b)^2=r^2[/tex]
Where M(a,b) and r:radius.
I understand this well, but when the subject is arcs...
[tex] (x-a)^2=r^2-(y-b)^2[/tex]
[tex] x_\textrm{1,2} =a (+-) \sqrt{r^2-(y-b)^2}[/tex]
My teacher said that equations for x1 and x2 were half circles at right and left. But how?
And also the same fo y,
[tex] y_\textrm{1,2}=b(+-)\sqrt{r^2-(x-a)^2}[/tex]
were the arcs of top half and bottom of the circle. But why?
Any help is appreciated.
 
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You apparently accept that [tex](x-a)^2+(y-b)^2=r^2[/tex]
is the equation of a circle.

To get [tex]x_\textrm{1,2} =a (+-) \sqrt{r^2-(y-b)^2}[/tex], you solve for x. Of course, with the square root, you have to take + and - to get both roots.

You know, I hope, that x measures right and left on a graph. The point (4,3) is 4 units to the right of the x-axis. The point (-4,3) is 4 units to the left. When you solve any equation for x, the result is "left" or "right". Taking the positive sign is right, negative, left.

y measures up and down so solving for y does the same thing except up and down instead of right and left.
 
I knew that basics but couldn't put together. Thanks anyway.
 

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