Find the equation of the diameter of the circle having equation $x^2+y^2-6x+2y-8=0$

Question

Find the equation of the diameter of the circle having equation $x^2+y^2-6x+2y-8=0$.

My Work: $$x^2+y^2-6x+2y-8=0$$ Comparing the above equation with $x^2+y^2+2gx+2fy+c=0$, we get: $$g=-3$$, $$f=1$$ Now, $$\textrm {centre}=(-g,-f)$$ $$=(3,-1)$$

Answer 1

Complete the square

$$x^2 + y^2 -6x + 2y -8 = 0$$ $$x^2-6x + 9 - 9 +y^2+2y + 1-1-8= 0$$ $$(x-3)^2+(y+1)^2 = 18$$

Therefore, $18 = r^2,$ and diameter $= 6\sqrt 2$.

Answer 2

Search for "Completing the Square".

Solving your system $x^{2}+y^{2}−6x+2y−8=0$, you can add and substract some numbers to get the classical circle equation $(x-h)^{2}+(y-k)^{2}=r^{2}$.

Well, you can check that your problem is equivalent to $$(x-3)^2+(y+1)^2=18$$ So your radius is $\sqrt{18}=3\sqrt{2}$, and your center $(3,-1)$

I dont know what you mean about "the equation of the diamater" but if you have a point $(a,b)$ the equation that passes through the point and the diamenter is: $$y=\frac{b-3}{a+1}(x-a)+b$$

Answer 3

Supposed the equation describes a surface with height function:

$$g(x,y) = x^2+y^2-6x+2y-8 $$

The circle is the intersection $g(x,y)=0$

To find the center of the circle. The way I do it is by solving the following 2×2 system $$\left. \begin{align} \frac{\partial g(x,y)}{\partial x}=0 \\\frac{\partial g(x,y)}{\partial y}=0 \end{align} \right\} \begin{aligned} x & = 3 \\ y & =-1 \end{aligned}$$

To find a line through the center $(3,-1)$ find the coefficients that solve $a x + b y + c =0$ with the above values. If you realize that $(a,b)$ is a vector perpendicular to the line then you can write $(a,b)=(-\sin \varphi,\cos \varphi)$ with $\varphi$ the angle of the line from the horizontal.

Now we solve $(-\sin \varphi) (3) + (\cos \varphi) (-1) + c = 0$ for $c=\cos\varphi+3 \sin\varphi$. So the equation of the line through the center of the circle is

$$ (-\sin\varphi) x + (\cos\varphi) y + (\cos\varphi+3 \sin\varphi) =0 $$

This can be simplified as

$$\boxed{ -(x-3) \sin\varphi + (y+1) \cos\varphi =0 }$$

Answer 4

Ok, so the center of that circle is: $(3,-1)$.

Now as there no other particular point is specified on this circle we can draw infinite number of diameters of that circle. Therefore there can be infinite number of straight line equations.

As for example the equation of the diameter of this circle which passes through the origin can be given by: $$\dfrac{y+1}{x-3}=\dfrac{-1-0}{3-0}\\ \implies x+3y=0$$