Find the radius and centre of the circle $x^2 -6x +y^2 -2y -6=0$
Can someone please help me with this question? I'm quite lost with what I have to do.
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There are much easier approaches, but I think it's always good to start by understanding what we're supposed to be doing. If you recognize the equation as describing a circle in the plane, and need to locate its center, then what you're doing is looking for a shift, that is $(x,y)\mapsto (u,v)=(x+a,y+b)$, such that the equation becomes that of a circle about the origin: $u^2+v^2=r^2$. So, you can substitute $x$ for $u-a$ and $y$ for $v-b$, and then try to find $a,b$ that would get rid of all the unwanted terms. – Jonathan Y. Oct 16 '13 at 11:00
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HINT Try to complete the square $x^2-6x$ to get it into the form of $(x-a)^2-c$, do the same for $y$ and collect the numbers on the RHS.
draks ...
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Sorry, quick question. Do I need to separate the x and y of the original equation on RHS and LHS first? – Red Queen10101 Oct 16 '13 at 11:05
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no keep'em on the LHS, but in fact it doesn't matter... just complete the squares...maybe you like to put some brackets $(x^2-6x +c-c)=(x-a)^2-c$ – draks ... Oct 16 '13 at 11:07
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You need to write $x^2-6x+y^2-2y-6=0$ as $(x-x_0)^2+(y-y_0)^2=r^2$, for some $x_0, y_0, r$. Once you do this the center will be $(x_0, y_0)$ and the radius will be $r$.
Spoiler:
Try to prove that $x^2-6x+y^2-2y-6=0\iff (x-3)^2+(y-1)^2=6+9+1$.
antifb
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Hey, thanks for the reply to the question. Could you please explain how you got from the original equation to the equation of a circle. I just need a first look at how its done. Thanks for the help! – Red Queen10101 Oct 16 '13 at 11:02
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(x-3)^2+(y-1)^2=4^2 where the coordinates of the centre is (3,1) and radius is 4 This is by completing the square method
John T Mendy
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