Question

The equation for the circle is: x^2 + y^2 +14x+10y−7=0 .   What is the center of the circle?

Answer 1

we are given

$x^2+y^2 +14x+10y-7=0$

Firstly , we can complete x square and y square

we can rearrange it

$x^2+14x+y^2 +10y-7=0$

$x^2+14x+7^2+y^2 +10y+5^2-7=0+7^2+5^2$

$x^2+14x+7^2+y^2 +10y+5^2=0+7^2+5^2+7$

$(x+7)^2+(y+5)^2=0+7^2+5^2+7$

$(x+7)^2+(y+5)^2=81$

we can also write it as

$(x+7)^2+(y+5)^2=9^2$

now, we can compare it with standard equation of circle

$(x-h)^2+(y-k)^2=r^2$

so, we will get

$h=-7 , k=-5 , r=9$

so,

$center=(-7,-5)$

$radius=9$................Answer