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Question - Find the equation of the circle which passes through $(3, 3)$ and $(5, 7)$ and has its center on the line: a): $x-y=5$

My attempt: $$m=2, b=-3, y=2x-3$$

Then I solved equation and got $(-2,-7)$ (which I feel is odd)

$\sqrt{(3-(-2))^2+(3-(-7))^2}$

$R= 5\sqrt5$

$(x+2)^2+(y+7)^2=225$

I know this is probably wrong but can anyone help me find my mistake.

CDXX
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    I’m voting to close this question because the person is not new to the site and has not yet started using mathjax. Post is unreadable. – Math Lover Jul 26 '21 at 12:18
  • ok do that .... – CDXX Jul 26 '21 at 12:19
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    Just in case nobody shared with you before, here is the link that helps you write math on the site https://math.meta.stackexchange.com/questions/5020/mathjax-basic-tutorial-and-quick-reference?noredirect=1&lq=1 – Math Lover Jul 26 '21 at 12:20
  • As I cannot follow your post, here are my suggestions on how you can go about solving it. There are multiple ways but here is couple of them - the center of the circle is $ O (x_0, x_0 - 5)$. Equate the square of distance from $O$ to the two given points and that will give you $x_0$ and hence the center and then the radius. Second, you can take midpoint $M$ of the given points and slope of $OM$ multiplied by $m = 2$ should give you $-1$ as they are perpendicular. – Math Lover Jul 26 '21 at 12:34
  • Does this answer your question? https://www.toppr.com/ask/question/find-the-equation-of-the-circle-passing-through-the-points-5537-and-has-its-center/ –  Jul 26 '21 at 12:35
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    I did some more formatting. Please check. Your mistake is that you are assuming that the center of the circle is at the intersection of both lines. That is not a correct assumption. The center is anywhere on line $x - y = 5$. What is the distance you get to point $(5, 7)$ from $(-2, -7)$? Not $5 \sqrt5$, right? Then how is it the center? – Math Lover Jul 26 '21 at 12:42
  • Yes u r right i got: $(x-8)^2+(y-3)^2=25$ – CDXX Jul 26 '21 at 12:46

1 Answers1

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Hint:
The center must be on the perpendicular bisector of $(3,3), (5,7)$ which is a line segment that passes through $(4,5)$.

Further, since the slope of the line that passes through the points $(3,3), (5,7)$ is $(2)$, the slope of the perpendicular bisector must be $(-1/2)$. Therefore, you have the slope of the perpendicular bisector, and one of its points, namely the point $(4,5)$.

This implies that you can deduce the equation that represents the perpendicular bisector. You also have the equation of the second line. Therefore, you can find the intersection point of the two lines.

user2661923
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