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

Question

hba · Answer

@waterineyes

nurali · Answer

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

hba · Answer

@waterineyes I got x^2+y^2+6x-10y-15=0

anonymous · Answer

What is center??

hba · Answer

Note that i used the formula (x-a)^2+(y-b)^2=r^2 @waterineyes

anonymous · Answer

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

hba · Answer

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

anonymous · Answer

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

hba · Answer

I am right :D

anonymous · Answer

Shouldn't it be :

$(x+3)^2 + (y-5)^2 = 7^2$ ??

hba · Answer

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

hba · Answer

@waterineyes And  I got x^2+y^2+6x-10y-15=0

anonymous · Answer

Explain to me what are you doing..

hba · Answer

@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

anonymous · Answer

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

anonymous · Answer

That was of Nurali..

hba · Answer

Yeah lol :D Thanks anyway :)

anonymous · Answer

Yes you are right..