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hba

  • 3 years ago

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

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  1. hba
    • 3 years ago
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    @waterineyes

  2. Nurali
    • 3 years ago
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    the standard equation of a circle: (x-h)^2 + (y-k)^2 = r^2 given r=7 c(5,-3) (x-5)^2+(y-(-3))^2=7^2 (x-5)^2+(y+3)^2=49 answer

  3. hba
    • 3 years ago
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    @waterineyes I got x^2+y^2+6x-10y-15=0

  4. waterineyes
    • 3 years ago
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    What is center??

  5. hba
    • 3 years ago
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    Note that i used the formula (x-a)^2+(y-b)^2=r^2 @waterineyes

  6. waterineyes
    • 3 years ago
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    Center is (-3,5) and you are using (5,-3) Any special reason??

  7. hba
    • 3 years ago
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    The center (a,b)=(-3,5) and r=7

  8. waterineyes
    • 3 years ago
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    Forget the formula.. Check once again for your solution..

  9. hba
    • 3 years ago
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    I am right :D

  10. waterineyes
    • 3 years ago
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    Shouldn't it be : \((x+3)^2 + (y-5)^2 = 7^2\) ??

  11. hba
    • 3 years ago
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    I am not using Nuralis Formula Dude i have solved it right :D

  12. hba
    • 3 years ago
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    @waterineyes And I got x^2+y^2+6x-10y-15=0

  13. waterineyes
    • 3 years ago
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    Explain to me what are you doing..

  14. hba
    • 3 years ago
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    @waterineyes (x+3)^2+(y-5)^2=7^2 x^2+6x+9+y^2-10y+25=49 x^2+y^2+6x-10y-15=0

  15. waterineyes
    • 3 years ago
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    Oh my God.. Sorry, I was thinking that, that was your post.. Really sorry..

  16. waterineyes
    • 3 years ago
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    That was of Nurali..

  17. hba
    • 3 years ago
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    Yeah lol :D Thanks anyway :)

  18. waterineyes
    • 3 years ago
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    Yes you are right..

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