Question

X^2+y^2+14x+10y=7 center and radius

Answer 1

$(x-a)^2+(y-b)^2=r^2\\\\(a;\ b)-the\ coordinates\ of\ the\ center\\r-the\ radius\\\\==================================\\Use:(k+l)^2=k^2+2kl+l^2\ (*)\\==================================$

$x^2+y^2+14x+10y=7\\\\x^2+14x+y^2+10y=7\\\\x^2+2x\cdot7+y^2+2y\cdot5=7\\\\\underbrace{x^2+2x\cdot7+7^2}_{Use\ (*)}-7^2+\underbrace{y^2+2y\cdot5+5^2}_{Use\ (*)}-5^2=7\\\\(x+7)^2-49+(y+5)^2-25=7\\\\(x+7)^2+(y+5)^2-74=7\ \ \ \ |add\ 74\ to\both\ sides\\\\(x+7)^2+(y+5)^2=81\\\\(x+7)^2+(y+5)^2=9^2$

$Answer:\\\boxed{(-7;-5)-center;\ 9-radius}$

Answer 2

$(x-a)^2+(y-b)^2=r^2$

$x^2+y^2+14x+10y=7\\ x^2+14x+49+y^2+10y+25-74=7\\ (x+7)^2+(y+5)^2=81\\ a=-7,b=-5, r^2=81\Rightarrow \hbox{center}=(-7,-5), r=\sqrt{81}=9$