Coordinate geometry -HELP PLEASE

[DBashev]

Find the centre and radius of the circle with equation X²+Y²-6X-2Y-6=0
I have some more but I need just the method , so please help me.

 

[Ishuda]

Find the centre and radius of the circle with equation X²+Y²-6X-2Y-6=0
I have some more but I need just the method , so please help me.

Lets do the general solution and see if that is enough:
Suppose we have an equation
a x² + a y² + bx + c y + d = 0
where a is not zero and b, c, and d are real valued. Well if a is not zero, we can divide through by a and get
x² + y² + bx + c y + d = 0
where I have just relabeled the other constants, i.e. replaced b/a with b, etc.

We now need to complete the squares for the x and y parts. We'll do x and y can be done the same way:
x² + bx = x² + 2 (b/2) x + (b/2)² - (b/2)² = (x + b/2)² - (b/2)²

Doing the same for y we can now write our original equation as
x² + y² + bx + c y + d = (x + b/2)² - (b/2)² + (y + c/2)² - (c/2)² + d = 0
or
(x + b/2)² + (y + c/2)² = (b/2)² + (c/2)² - d
Letting
r = $\displaystyle \frac{1}{2}\sqrt{b^2 + c^2 - 4 d}$
we finally have the equation of a circle [IF r is real, i.e. $\displaystyle b^2 + c^2 \ge 4 d$],
(x - x₀)² + (y - y₀)² = r²
with center at (x₀, y₀) and radius r where
x₀ = -b/2
y₀ = -c/2

As an aside, if $\displaystyle b^2 + c^2 \lt 4 d$ we would still have a circle but x and y would be complex numbers with real parts x₀ and y₀ and imaginary parts satisfying a circle of radius [EDIT: magnitude of] r centered (0,0) for the imaginary part.

 

[stapel]

Find the centre and radius of the circle with equation X²+Y²-6X-2Y-6=0
I have some more but I need just the method....

There are loads of online resources showing the method. Try this listing. Once you have studied at least two lessons from the list, please attempt the exercise. If you get stuck, you can then reply with a clear listing of your efforts so far.

Thank you!