what is the center and radius of the circle with the given equation? x^2+y^2+14x-12y=-69
so do you need the exact number or could you have x and y's?
it would help me better if i had the numbers i would think
uhh well you need 2 equations to solve for this mess and you now only have 1 equation which means that variable x and variable y can be some unfixed number
oh okayy
so do you have a second equation?
on my study guide this is the question has its written out i just need some help
you need to convert the equation to the form (x - a)^2 + (y - b)^2 = r^2 this will have centre (a,b) and radius r
oh yea what galli said
hmm skellz do you know what a "perfect square trinomial" is?
first convert x^2 + 14 to the form (x + c)^2 - d and do similar coversion with y^2 - 12y
x^2 + 14x = (x + 7)^2 - 49
you have the 14 and then subtract the square of this half
ok i went to the bathroom let me catch up with what yall are saying but no i dont know that that is
ur formula comes down to \[(x^2+14x+49)-49+(y^2-12y+36)-36\]=-69
yes
move all the things over then you can factor into perfect squares so you can find the radius of your circle through \[x^2+y^2=r^2\]
the perfect square is in the parentheses
^
can you finish this now? or do u need more help
ok omm
i got an answer but not sure if its right
what is it?
center at (7,-6) radius 16
your final equation should be ( x + 7)^2 + ( x + 6)^2 = 85 - 69 = 16 not quite 7 = -a ( see the standard equation) and 6 = -b) , 16 = r^2 so a = -7, b = -6 and r = 4
ohhhhh
get it?
appreciate it
yw
oh - i made an error sorry its (y - 6) ^2 not (x+ 6(^2 - as its -12y so b = 6 correct is center( -7,6) and radius 4
* not (y + 6)^2 as its -12y
i got the sign wrong
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