SOLUTION: One pair of integer values satisfy the equation x^2+y^2=100 is x=6 and y=8. Find pair of integer values that satisfy x^2+6x+y^2-4y-87=0. Thanks in advance!

Question 1000937: One pair of integer values satisfy the equation x^2+y^2=100 is x=6 and y=8. Find pair of integer values that satisfy x^2+6x+y^2-4y-87=0. Thanks in advance!
Found 2 solutions by stanbon, Cromlix:
Answer by stanbon(75887) (Show Source): You can put this solution on YOUR website!
One pair of integer values satisfy the equation x^2+y^2=100 is x=6 and y=8. Find pair of integer values that satisfy x^2+6x+y^2-4y-87=0. Thanks in advance!
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Substitute for x^2 + y^2 to get:
6x - 4y + 100 - 87 = 0
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6x - 4y + 13 = 0
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y = (6x+13)/4
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Find integer value "x" that has corresponding integer value "y".
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Ans: There is no such pair of integers.
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Cheers,
Stan H.
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Answer by Cromlix(4381) (Show Source): You can put this solution on YOUR website!
Hi there,
With equation of a circle:
x^2 + y^2 + 6x - 4y - 87 = 0
The centre of the circle is (-3,2)
The radius = 10
(sqrt (-3)^2 + (2)^2 -(-87))
If you move along horizontally from
coordinates (-3,2) by 10 units the
set of coordinates (7,2) is reached.
This point is on the circle.
These will satisfy the equation:-
x^2 + y^2 + 6x - 4y - 87 =0
(7)^2 +(2)^2 + 6(7) - 4(2) - 87 = 0
49 + 4 + 42 - 8 - 87 = 0
Hope this helps :-)

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