How do you write ${(x - 17.09)}^{2} + {(y - 1.53)}^{2} = {1.03}^{2}$ $(x-17.09)^2+(y-1.53)^2=1.03^2$ into the general form?

Question

1747 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-10-13T19:06:59

This is the equation of a circle that is offset such that the centre is at

$(x, y) \to (17.09, 1.53)$ $(x,y)->(17.09,1.53)$ and it has a radius of length $1.03$ $1.03$

This basically is already in general form for a circle.

Standard form is ${(x - h)}^{2} + {(y - k)}^{2} = r^{2}$ $(x-h)^2+(y-k)^2=r^2$

Where $r$ $r$ is the radius and $(h, k)$ $(h,k)$ is the centre.

Tony B

How do you write (x−17.09)2+(y−1.53)2=1.032(x-17.09)^2+(y-1.53)^2=1.03^2 into the general form?