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I'm having trouble with some questions relating to finding the centre coordinates and radius of a circle when given the equation. I understand how to find it in the form: (x-p)^2 + (y-q)^2 = r^2 I.e centre (p,q), radius= r

What confuses me is when for example the question is: 25x^2 + 25^y^2 = 9 In this case i would presume the answer is centre(0,0), radius = 3 However the answer is actually centre(0,0), radius= 0.6.

I presume it's something to do with the 25 but i don't see how.

Another example: x^2 + y^2 - 6x + 4y + 4 = 0 I have no idea how to work out the centre. But i thought the radius should equal root of -2, but that is not a real number...

Anyway, any help would be much appreciated. Thanks :)

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  • $\begingroup$ For your first question, divide both sides of your equation by $25$ to get $x^2+y^2=\frac{9}{25}$ and you'll see it. For your second question, complete the square. $(x-3)^2-9+(y+2)^2-4+4=0$. $\endgroup$ – Michael Burr May 6 '17 at 13:14
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$$x^2+y^2-6x+4y+4=0$$

Perform completing the squares:

$$(x^2-6x)+(y^2+4y)+4=0$$

$$(x^2-6x+9)-9+(y^2+4y+4)-4+4=0$$

$$(x-3)^2-9+(y+2)^2=0$$

$$(x-3)^2+(y+2)^2=9=3^2$$

Remark: If the coefficient for $x^2$ is not $1$, you might want to divide the equation using that number first.

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