[ASK] Equation of a Circle

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Discussion Overview

The discussion revolves around finding the standard equations of circles with centers on the line defined by the equation 4x + 3y = 8, which are also tangent to the lines x + y = -2 and 7x - y = -6. The scope includes mathematical reasoning and problem-solving related to geometry and algebra.

Discussion Character

  • Mathematical reasoning
  • Exploratory
  • Homework-related

Main Points Raised

  • One participant presents initial equations involving parameters a and b, suggesting a relationship between the center of the circle and the tangential conditions.
  • Another participant questions the relevance of the parameters a and b, proposing that the center (a, b) should be defined explicitly and relates it to the line equation.
  • A method is suggested to derive the conditions for tangency by forming a quadratic equation based on the circle's equation and the tangential lines.
  • A later reply expresses relief upon realizing that the problem indeed allows for two possible circle centers, indicating a potential misunderstanding or concern about the uniqueness of the solution.

Areas of Agreement / Disagreement

Participants do not reach a consensus on the approach to solving the problem, as there are differing interpretations of the initial equations and the implications of the tangency conditions. The discussion remains unresolved regarding the final formulation of the circle equations.

Contextual Notes

There are limitations in the clarity of the initial equations presented, as well as potential dependencies on the definitions of parameters used. The discussion also reflects uncertainty about the uniqueness of the solutions for the circle centers.

Monoxdifly
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find standard eqations of circles that have centers on 4x+3y=8 and are tangent to both the line x+y=-2 and 7x-y=-6

What I got is $$4a=–4\pm3r\sqrt2$$ and $$b=4\pm r\sqrt2$$. Dunno how to continue from here.
 
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Monoxdifly said:
find standard eqations of circles that have centers on 4x+3y=8 and are tangent to both the line x+y=-2 and 7x-y=-6

What I got is $$4a=–4\pm3r\sqrt2$$ and $$b=4\pm r\sqrt2$$. Dunno how to continue from here.
I'm afraid this makes no sense because there are no "a" or "b" in the original question so we have no idea how this relates to the question.

Here's what I would do:
Let (a, b) be the center of such a circle. (I think that's what you intended but did not say.) Then, since the center lies on the line 4x+ 3y= 8, we have b= (8- 4a)/3. The circle can be written (x- a)^2+ (y- (8-4a)/3)^2= r^2 and we need to find values for a and r. We have two more conditions.

If that circle crossed the line x+ y= 2 then the equation (x- a)^2+ (2- x- (8- 4a/3)^2= r^2 would have 2 solutions. But if the line is tangent to the circle, that quadratic equation must have a double root. Do the indicated squares in
(x- a)^2+ (2- x- (8- 4a/3)^2= r^2 to get the quadratic equation in standard form and use the quadratic formula to see what must be true about a and b.


Do the same with the line 7x- 4y= -6.
 
Wow, didn't think that the steps were far simpler than I thought, though I still needed half a day to arrive at the final answer due to my lack of accuracy during calculating some things. Thanks for your help.
 
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I freaked out yesterday when my calculation got two centers of circle. Now I'm relieved that it does have two circles.
 

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