SOLUTION: What is the sum of the coordinates of the center of the circle whose equation is x^2 + y^2 - 6x - 10y + 30 = 0 ?

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Question 366780: What is the sum of the coordinates of the center of the circle whose equation is
x^2 + y^2 - 6x - 10y + 30 = 0 ?
Found 3 solutions by stanbon, mananth, ewatrrr:
Answer by stanbon(75887)   (Show Source): You can put this solution on YOUR website!
)What is the sum of the coordinates of the center of the circle whose equation is
x^2 + y^2 - 6x - 10y + 30 = 0 ?
----
x^2 - 6x + 9 + y^2-10y+25 = -30+9+25
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(x-3)^2 + (y-5)^2 = 4
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Center: (3,5)
Sum = 8
==============
Cheers,
Stan H.
Answer by mananth(16946)   (Show Source): You can put this solution on YOUR website!
x^2 + y^2 - 6x - 10y + 30 = 0
..
x^2-6x +y^2-10y+30=0
(x^2-6x+9)-9+(y^2-10y+25)-25+30=0
(x-3)^2+(y-5)^2-9-25+30=0
(x-3)^2+(y-5)^2=9+25-30
(x-3)^2+(y-5)^2=4
Center = (3,5)
sum = 8
...
m.ananth@hotmail.ca
Answer by ewatrrr(24785)   (Show Source): You can put this solution on YOUR website!


Hi, 

The standard equation of a circle with center C(h,k) and radius r is :



x^2 + y^2 - 6x - 10y + 30 = 0

x^2 -6x + 9 + y^2 -10y +25 + 30 -9 -25 = 0

(x^2 -6x + 9) + (y^2 -10y +25) + (30 -9 -25) = 0

(x-3)^2 + (y-5)^2 -4 = 0

(x-3)^2 + (y-5)^2 = 2^2

sum of the coordinates(3,5) of the center of the circle

3 + 5 = 8 

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