Question

wow, super long problem...got all the way down to th ebottom thinking i had this one right but it was still wrong. can any! pls help? i wonder where i went wrong w/ this

1) find the standard form of the equation of the circle w/ endpoints of a diameter at the points (1,4) and (-5,2)

***TYPE standard form of the equation of the circle.

Answer 1

d=2r
soo d/2 =r

Answer 2

half the distance between points is the radius
midpoint is the center

Answer 3

using distance formula you should have gotten the the diameter should be

d=sqrt(40)

r=sqrt(10)

Answer 4

indeed so you'll have to use the midpoint formula and then the distance formula

Answer 5

or you could simply find the distance between the 2 pts and divide by two

Answer 6

either way you need your center which is given by the midpoint

Answer 7

so long to type but i can show what i did towards the end: center then was (-2,3) and i found dist. to either point (1,4) or (-5,2)...anyway i ended up w/ a radius of 28
(x-h)^2+(y-k)^2=r^2
(x-(-2))^2+(y-3)^2=(sqrt28)^2
(x-2)^2+(y-3)^2=28 AND I WAS WRONG! ;(

Answer 8

anyways looks like you have a lawge amount of people here i'm sure they can figure it out

Answer 9

Look the radius is \[\sqrt{40}/2\]

Answer 10

YOU GUYS ARE SO QUICK, I'M AMAZED! thank you outkast, dumbcow, dichiaraj and vicky...wow 40 then? i'm so afr off. if that's the case how would i thne write the last part of this equation? something like (y-2)^2+(y-3)^2=______???

Answer 11

well \[\sqrt{40}/2\] -- can also be written as

\[\sqrt{4*10}/2= 2*\sqrt{10}/2= \sqrt{10}\]

so that is the radius and since the equation asks for radius squared it should just be 10

Answer 12

hmm
\[d = \sqrt{(x2-x1)^{2} + (y2-y1)^{2}}\]
\[d = \sqrt{(-5-1)^{2} + (4-2)^{2}} = \sqrt{36+4} = \sqrt{40} = \sqrt{4*10} = 2\sqrt{10}\]

Answer 13

also if you center is at (-2,3) your equation should be

\[(x+2)^2+(y-3)^2=10\]

no mine is x+2 yours you put down was x-2 it should be x+2

Answer 14

lol dichiaraj...i got mine the other way around. i'm impressed at how you you ALL solve problems...seriously...there's just so many ways.thank you guys.