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

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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

User Ohad Horesh
by
8.2k points

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