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.