Another conics question. (Need tutored, no idea about conics).
Find the center and radius of a circle that has a diameter with endpoints(-9,-6) and (-1,0).
hmm here we go bty you can google "conics purplemath" and get real good explanations
oh i will google that too c:
ok first you need the center of the circle, which is another way of saying you need to find the midpoint of the line segment
Isnt it (-9)+1 / (-6)-0?
you need help with that, or is it clear how to find the midpoint?
oh no, i think you have the midpoint confused with the "slope" which has nothing to do with this problem
Oh yeah.. I cant remember the midpoint formula /:
the midpoint is just the average of each coordinate
i.e. add up and divide by 2
add up x and x? or x and y?
\[\frac{-9-1}{2}=\frac{-10}{2}=-5\] for the \(x\)
so -5,-3?
\(x\) and \(x\) for the \(x\) \(y\) and \(y\) for the \(y\) you cannot add or subtract \(x\) and \(y\) they live in different yes
now we area half way done the general form is \[(x-h)^2+(y-k)^2=r^2\] and you have \(h=-5, k=-3\)
and the radius is half of the diameter, so (-5,-3) to -1,0) is the radius, so I have to use the distance formula?
so far we know the equation is going to look like \[(x+5)^2+(y+3)^2=r^2\]
yes exactly you have to use the distance formula to find \(r\) but before you do, notice that you really need \(r^2\) and not \(r\) so you can use the distance formula without the square root
in other words compute \[(-1+5)^2+3^2\]
25
i hope that is clear, that will give you \(r^2\) which is what you want i get \(4^2+3^2=25\)
right
now you are done
But is the radius 25 units or 5 units?
\[(x+5)^2+(y+3)^2=25\]
the radius is \(r\) and since \(r^2=25\) you know \(r=5\)
Makes perfect sense! Thank you again c:
yw
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Haha c:
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