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 :)
$\endgroup$ 11 Answer
$\begingroup$$$x^2+y^2-6x+4y+4=0$$
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.
$\endgroup$ 0
