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$A=(3,1)$ and $B=(-1,-1)$ are points on a circle of center $(k, -3k)$ find the value of $k$

I begin by assinging the values $\ g = -k $ and $\ f=3k $.

I then substitute $(3, 1)$ and $ g= -k, f= 3k $ into $\ x^2+y^2+2gx+2fy+c$. That leaves me with $c = 10$, meaning $k= -2$. Where have I gone wrong? I know its wrong.

FINAL EDIT:

Very sorry, as foolish as this sounds, there is a mix up in the numbering of the answers, I actually got the answer correct. Thank you for your time, sorry about wasting it.

Please delete this Question.

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    $\begingroup$ I don't understand what you do. Why substituting $(3,1)$ in the equation leaves you with $c=10$? How can you get to an answer without ever using the point $B$? $\endgroup$ – Andrea Mori Jan 31 '12 at 23:54
  • $\begingroup$ $10 + 2(-k)(3) + 2(3k)(1) + c$ makes $10 -6k +6k + c$ that means $c=10$ $\endgroup$ – Rollo Montgomery Konig-Brock Feb 1 '12 at 0:00
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    $\begingroup$ @Rollo: Hi! Hope you've been enjoying math.SE so far. Just wanted to explain that I don't think there's a need to delete; while you've determined what your issue was, there might be someone else with a similar question who will find the answers here helpful, and we'd want the site to be able to useful to them too. $\endgroup$ – Zev Chonoles Feb 1 '12 at 3:54
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I'd approach the problem differently. First, the perpendicular bisector of any chord of a circle contains the center of the circle, so I'd start by writing an equation for the perpendicular bisector of $\overline{AB}$. Then, substitute $(k,-3k)$ into that equation and solve for $k$.

Using this method, I also get $k=-2$; how do you know that is wrong?

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  • $\begingroup$ $x^2 + y^2 - 10x - 4y +4$ is the equation of the circle, this means that c=4 and the center is not (-2, 6). EDIT: really sorry I know I'm being tiresome. Sorry mods. $\endgroup$ – Rollo Montgomery Konig-Brock Jan 31 '12 at 23:51
  • $\begingroup$ @RolloMontgomeryKonig-Brock: That can't be the equation of the circle—the center of the circle with that equation is at $(5,2)$, which is not of the form $(k,-3k)$. (Also, no equals sign, so it's not an equation.) $\endgroup$ – Isaac Jan 31 '12 at 23:53
  • $\begingroup$ I edited the equation, sorry $\endgroup$ – Rollo Montgomery Konig-Brock Jan 31 '12 at 23:55
  • $\begingroup$ @RolloMontgomeryKonig-Brock: Where did you edit the equation? In your first comment, there's still no equation (nothing with $x$ and $y$ and an = in it); if the first expression is intended to be an equation (for example, with $=0$ at the end), it still is a circle centered at $(5,2)$, which isn't of the form $(k,-3k)$. $\endgroup$ – Isaac Jan 31 '12 at 23:57
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I don't know what that $K$ is and I don't care as I don't need to know it for my solution.

Issac has given an answer appealing to geometry. That you mention the equation of circle and seem to be working with it, this might be of some help as well:

So, the equation of the circle becomes, $$x^2+y^2-2kx+6ky+c=0$$

Now plugging in $(3,1)$, you'll have, $$9+1 -6k+6k+c=0 \implies c=-10$$

Note the negative sign here.

Now, plug in $(-1,-1)$, you'll have, $$1+1+2k-6k-10=0 \implies-4k-8=0 \implies k=-2$$

Hence, $\boxed{c=-10;k=-2}$.

Hope this helps.

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Here's another variation. There will be a circle centered at $C=(k,3k)$ and going through $A$ and $B$ if and only if $A$ and $B$ are equidistant from $C$. Analitically, if and only if $$ (3-k)^2+(1+3k)^2=(-1-k)^2+(-1+3k)^2. $$ The equation simplifies to $8+4k=0$, i.e. $k=-2$.

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Since ( k, -3k ) is the centre of the circle and the radius(r) of the circle is equal to r^2 = (k-3)^2 + (-3k-1)^2 and, r^2 = (k+1)^2 + (-3k+1)^2 ... by equating the two equations we have k=-2

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    $\begingroup$ Does this answer add anything new to the existing answers (especially the one by Andrea Mori)? $\endgroup$ – Joonas Ilmavirta Nov 13 '14 at 16:18

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