Finding coordinates of a point on a circle( angle and distance from O known)

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SUMMARY

The discussion focuses on finding the coordinates of point P on a circle, given the angle and distance from the center O. The equation of the circle is defined as (x-a)² + (y-b)² = r², where (a, b) is the center and r is the radius. Participants discuss substituting the line equation y = mx + c into the circle's equation to find the intersection points. The conclusion emphasizes that point P can vary along the arc of the circle, and the correct substitution is crucial for solving the quadratic equation derived from the intersection of the line and the circle.

PREREQUISITES
  • Understanding of circle equations, specifically (x-a)² + (y-b)² = r²
  • Knowledge of linear equations in the form y = mx + c
  • Familiarity with solving quadratic equations
  • Basic geometry concepts related to circles and lines
NEXT STEPS
  • Learn how to derive the intersection points of a line and a circle using algebraic methods
  • Study the properties of quadratic equations and their solutions
  • Explore graphical methods for visualizing the intersection of lines and circles
  • Investigate the implications of varying the gradient m in line equations
USEFUL FOR

Mathematicians, physics students, and anyone involved in geometry or computational geometry who needs to understand the relationship between lines and circles.

Wikeda
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zn0sq9.jpg



1. I basically have to find the coordinates of P. All the pink lines are know, coordinates of points A and centre of circle are know.


2. (x-a)^2 + (y-b)^2 = r^2


3. I try to substitute mx+c into the equation and get

(x-a)^2 + (y-mx-c)^2 + r^2= 0

but I can't work out what m and c are. Any help would be appreciated! Am I in the right direction, or am I completely off?
 
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Wikeda said:
zn0sq9.jpg
1. I basically have to find the coordinates of P. All the pink lines are know, coordinates of points A and centre of circle are know.2. (x-a)^2 + (y-b)^2 = r^23. I try to substitute mx+c into the equation and get

(x-a)^2 + (y-mx-c)^2 + r^2= 0

but I can't work out what m and c are. Any help would be appreciated! Am I in the right direction, or am I completely off?


I'm just going off what it looks like on the picture, but it looks as though P is the intersection between a line with gradient -1 going through the centre of the circle and the circle itself. Is this correct?

If so, what is the equation of a line with gradient m that goes through the point (a,b) ?

EDIT: And by the way, if you were trying to find where a general circle intersects with a general line, then you need to plug
y=mx+c into the equation
(x-a)^2+(y-b)^2=r^2
to get
(x-a)^2+(mx+c-b)^2=r^2
And then you'll have an equation that you have a quadratic in x that you have to solve (all the other values are constants, so the answer will depend on what those are).
Once you've done that, you can plug that value of x back into the line equation to find y (you can also plug into the circle equation but it's harder work to solve for y there and you also get two values).
 
Thank you for your response!
Actually The point P could be anywhere on the arch that is being formed by the circle, so the line doesn't intersects the centre of the circle.

I see now where I got it wrong, should have substituted y instead of b.

Thanks!
 
Wikeda said:
Thank you for your response!
Actually The point P could be anywhere on the arch that is being formed by the circle, so the line doesn't intersects the centre of the circle.

I see now where I got it wrong, should have substituted y instead of b.

Thanks!

Well then, if you knew the equation of the circle, you won't know P exactly because it obviously will change depending on how high up the circle it is. The equations are still correct though :smile:
 

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