MHB [ASK] Equation of a Circle

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To find the standard equations of circles with centers on the line 4x + 3y = 8 and tangent to the lines x + y = -2 and 7x - y = -6, the center can be represented as (a, b), where b is expressed in terms of a. The equations of the circles can be formed using the center coordinates and the radius r, leading to the form (x - a)² + (y - (8 - 4a)/3)² = r². The tangency conditions require that the resulting quadratic equations from substituting the line equations into the circle equation must have double roots. The discussion concludes with the realization that there are indeed two valid circle centers, alleviating initial concerns about the calculations.
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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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