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 :)

  • $\begingroup$ 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


Perform completing the squares:





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

| cite | improve this answer | |

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.