hba
Equation of a circle with center (3,5) and radius 7 is :



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hba
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@waterineyes

Nurali
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the standard equation of a circle:
(xh)^2 + (yk)^2 = r^2
given r=7 c(5,3)
(x5)^2+(y(3))^2=7^2
(x5)^2+(y+3)^2=49 answer

hba
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@waterineyes I got x^2+y^2+6x10y15=0

waterineyes
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What is center??

hba
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Note that i used the formula (xa)^2+(yb)^2=r^2 @waterineyes

waterineyes
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Center is (3,5) and you are using (5,3)
Any special reason??

hba
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The center (a,b)=(3,5) and r=7

waterineyes
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Forget the formula..
Check once again for your solution..

hba
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I am right :D

waterineyes
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Shouldn't it be :
\((x+3)^2 + (y5)^2 = 7^2\) ??

hba
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I am not using Nuralis Formula Dude i have solved it right :D

hba
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@waterineyes And I got x^2+y^2+6x10y15=0

waterineyes
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Explain to me what are you doing..

hba
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@waterineyes (x+3)^2+(y5)^2=7^2
x^2+6x+9+y^210y+25=49
x^2+y^2+6x10y15=0

waterineyes
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Oh my God..
Sorry, I was thinking that, that was your post..
Really sorry..

waterineyes
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That was of Nurali..

hba
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Yeah lol :D Thanks anyway :)

waterineyes
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Yes you are right..