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.
[Math] Find the radius and centre of the circle $x^2 -6x +y^2 -2y -6=0$
circles
Related Solutions
The equation of a circle with centre $(a,b)$ and radius $r$ is $(x-a)^2+(y-b)^2=r^2$.
So you know the distance $d=CD$ between center and boundary. Then you can write
$$\cos\angle ECD = \tfrac dr \qquad \angle ECF = 2\angle ECD = 2\arccos\tfrac dr$$
Now the length of an arc is $r$ times its angle, so the outside arc is $r\angle ECF$ and the inside arc is
$$s = r(2\pi-\angle ECF)=2r(\pi-\arccos\tfrac dr)$$
You want that number to be equal to some given value, so you want to solve the above equation for $r$. Unfortunately, that equation is transcendental, so you can't expect a closed form solution to your problem. Your best bet is some form of iterative numeric approximation.
As you can see from that plot, you can expect that for many possible ratios of $\frac sd$, you get two distinct solutions for $r$.
The apex (with the vertical tangent) appears to be at
$$ s/d\approx 5.94338774142760424162091392488776998544210982523814509283191138267355981 \\ r/d\approx 1.06193134974748196175464922830803488867448733227482933642882008697882597 $$
Best Answer
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.