How do you decide whether or not the equation below has a circle as its graph If it does, give the center and the radius. If it does not, describe the graph `25x^2 +25y^2- 30x+ 30y -18 =0`?

The General Second Degree Equation in `RR^2`

`ax^2+2hxy+by^2+2gx+2fy+c=0` will represent a Circle , if,

`(i): a=b!=0, (ii): h=o, &, (iii): g^2+f^2-ac>0.`

In the event, its Centre is `(-g/a,-f/a)` and Radius is

`sqrt(g^2+f^2-ac)/|a|`.

In our Example, `(i): a=25=b!=0, (ii): h=0,` and,

`(iii): g=-15, f=15, c=-18`

`rArr g^2+f^2-ac=225+225=900>0`.

So, the eqn. represents a circle having centre #(15/25,-15/25), i.e.,

`(3/5,-3/5)" & radius "sqrt900/|25|=30/25=6/5`.

Enjoy Maths.!

Given: `25x^2+25y^2-30x+30y-18=0 ...............Eqn(1)`

If this an equation of a circle then we should be able to manipulate it back into standard form of `(x-a)^2+(y-b)^2=r^2`
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I spot that the standard form is part of the process for completing the square. Lets investigate that.

Divide throughout by 25 to get rid of the coefficients for `x^2 and y^2`

`x^2+y^2-6/5x+6/5y-18/25=0`

Write as:

`(x^2-6/5x)+(y^2+6/5y)=+18/25`

Completing the squares

`(x-6/10)^2-(-6/10)^2+(y+6/10)^2-(+6/10)^2=18/25`

`(x-3/5)^2+(y+3/5)^2=18/25+36/100+36/100 = 36/25`

Thus the centre as at point 1`->P_1->(x,y)=(3/5,-3/5)`

The radius is `r^2=36/25=> r = 6/5`

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`color(blue)("Putting it all together")`

`25x^2+25y^2-30x+30y-18=0`

Is the same as:

`(x-3/5)^2+(y+3/5)^2=r^2=(6/5)^2`

Thus it is a circle.