HELP ME WITH THIS PROBLEM PLZ
helpp
we can do this
it is kind of a pain, but totally doable first we need the point \((x,y)\) equidistant from the three points if it is equidistant from \((1,3)\) and \((5,11)\) then we know by the distance formula that \[(x-1)^2+(y-3)^2=(x-5)^2+(y-11)^2\]
okay
soo do i just plug in the points?
oh no!! we have a few more steps
multiply all that mess out and you get \[8 x+16 y-136 = 0\]
divide by 4 and get \[2x+4y=34\] now hold on to that, we have to do it again
ok
this time we use the third point \((11,4)\) and the distance formula gives \[(x-11)^2+(y-4)^2=(x-5)^2+(y-11)^2\]
multiply all that mess out, and you get \[-12 x+14 y = 9\]
Multiplying what exactly can you be more specific plz? sorry
that means we have to solve two equations \[x+2y=17\] and \[-12x+14y=9\]
oh i mean expand, and combine like terms
ooh ok
here it is for the second one \[(x-11)^2+(y-4)^2=(x-5)^2+(y-11)^2\] expanding gives \[x^2-22x+121+y^2-8yt+16=x^2-10x+25+y^2-22y+121\]
i got a typo there, but you get the idea get rid of the \(x^2\) and \(y^2\) from both sides add up etc
ok
hmm i wonder if i made a mistake, because my answer is very ugly hold on a sec
alright
no i got the same ugly answer twice, once using wolfram
http://www.wolframalpha.com/input/?i=%28x-1%29^2%2B%28y-3%29^2%3D%28x-5%29^2%2B%28y-11%29^2%2C%28x-5%29^2%2B%28y-11%29^2%3D%28x-11%29^2%2B%28y-4%29^2 \[x=\frac{110}{19},y=\frac{213}{38}\]
so this is the answer?
What other steps do i need to do?
now that we have the center of the circle (presumably the shape is a circle) then you have to find the equation for the circle so you need the distance from the center to any of the three other points they should all be the same
How do i find the equation??
\[(x-\frac{110}{19})^2+(y-\frac{213}{38})^2=r^2\] we need \(r^2\)
i can't believe they gave a problem with such an ugly answer
lol.. but how do find the radius? can set it up so i can figure it out
typo!! \[r^2=(\frac{110}{19}-1)^2+(3-\frac{213}{38})^2\]
Okay is there any other steps after this???
use a calculator if it was me, i would cheat
i mean cheat for the whole problem hold on
http://www.wolframalpha.com/input/?i=circle+through+%281%2C3%29%2C+%285%2C11%29%2C%2811%2C4%29
ok
Oh thank you :)
there we have it btw you are the second person lately to use myers briggs whats up with that??
I am? lol I didnt notice anyone else using it.
not yours, but some other one
infj
oh ok lol
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