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