Rewrite a circle's equation to easily see centre and radius

Question

$$x^{2}+y^{2}-5x-15y+30=0$$

I'm supposed to rewrite this equation so that you can easily see the centre and radius of the circle. I don't even know where to start. According to Mathematica the centre is $(5/2, 15/2)$ and the radius is $\sqrt{65/2}$.

Answer 1

Answer:

$$x^2 +y^2 -5x-15y+30 = 0$$ $$x^2 -5x +(\frac{5}{2})^2 +y^2 -15x +(\frac{15}{2})^2 -(\frac{5}{2})^2-(\frac{15}{2})^2+30 = 0$$

$$(x-\frac{5}{2})^2+(y-\frac{15}{2})^2 = \frac{225+25}{4}-30$$

$$(x-\frac{5}{2})^2+(y-\frac{15}{2})^2 =(\sqrt{\frac{65}{2}})^2$$

Center$$ (\frac{5}{2},\frac{15}{2})$$and the radius $$= \sqrt{\frac{65}{2}}$$

Answer 2

I will do $x$. Please do $y$.

$$x^2+y^2-5x-15y+30=0$$ $$x^2-5x+25/4-25/4+y^2-15y+30=0$$ $$(x-5/2)^2+y^2-15y+30-25/4=0$$

Answer 3

Hint :

$$\left(x^2-2\cdot\frac{5}{2}x+\frac{25}{4}\right)-\frac{25}{4}+\left(y^2-2\cdot\frac{15}{2}y+\frac{225}{4}\right)-\frac{225}{4}+30=0$$