Hello,
(\({x}^{2}-4x+{y}^{2}-6y-36=25\)),
(\(x^2-4x-y^2-6y=25+36\)),
(\(x^2-4x+y^2-6y=61\)),
(\(x^2-4x+?+y^2-6y=61+?\)),
to solve the equation \(x^2-4x+4=(x-2)^2\),
we are adding 4:
(\(x^2-4x+4+y^2-6y=61+4\)),
factorise the expression using square of a binomial (\(a^2-2ab+b^2=(a-b)^2\)),
\((x-2)^2+y^2-6y=61+4\)
\((x-2)^2+y^2-6y=65\),
\((x-2)^2+y^2-6y+?=65+?\),
to solve the equation \(y^2-6y+9=(y-3)^2\),
we are adding 9:
\((x-2)^2+y^2-6y+9=65+9\),
factorise the expression using square of a binomial \(a^2-2ab+b^2=(a-b)^2\),
\((x-2)^2+(y-3)^2=65+9\),
\((x-2)^2+(y-3)^2=74\),
The equation can be written in the form \((x-p)^2+(y-q)^2=r^2\)
so that it corresponds to a circle with radius (\(r=\sqrt{74}\)) and centre (\(2,3\)).
Straight
