What are the Intersection Points of a Circle and a Line?

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The discussion revolves around determining the intersection points between the circle defined by the equation x² + y² = 25 and the line represented by y = x + k, where k is a real number. The key approach involves substituting the line's equation into the circle's equation to form a quadratic equation in terms of x. The number of intersection points—one, two, or none—depends on the discriminant of this quadratic equation. The conversation highlights the importance of understanding how the discriminant indicates the nature of the solutions. Ultimately, the original poster found a solution to their problem.
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Homework Statement


I'm completely stuck on this question
I'll write the whole thing down

Consider the circle x² + y² = 25 and the line y = x + k. where k is any real number. Determine the values of k for which the line will intersect the circle in one, two or no points.

PLEEEASE help me
Thanks


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The Attempt at a Solution

 
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Didn't you ask a question almost exactly like this the other day? What have you tried so far?
 
Points, (x,y), at which two curves intersect are x, y values that satisfy both equations. You are trying to find x and y that satisfy both x2+ y2= 25 and y= x+ k. If y= x+ k then the first equation becomes x2+ (x+ k)2= 25. That's a quadratic equation for x. When does a quadratic equation have two distinct solutions?
 
nvm i got it yesterday lol
thanx anyways
 
I tried to combine those 2 formulas but it didn't work. I tried using another case where there are 2 red balls and 2 blue balls only so when combining the formula I got ##\frac{(4-1)!}{2!2!}=\frac{3}{2}## which does not make sense. Is there any formula to calculate cyclic permutation of identical objects or I have to do it by listing all the possibilities? Thanks
Essentially I just have this problem that I'm stuck on, on a sheet about complex numbers: Show that, for ##|r|<1,## $$1+r\cos(x)+r^2\cos(2x)+r^3\cos(3x)...=\frac{1-r\cos(x)}{1-2r\cos(x)+r^2}$$ My first thought was to express it as a geometric series, where the real part of the sum of the series would be the series you see above: $$1+re^{ix}+r^2e^{2ix}+r^3e^{3ix}...$$ The sum of this series is just: $$\frac{(re^{ix})^n-1}{re^{ix} - 1}$$ I'm having some trouble trying to figure out what to...
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