Question

Find the equation of the circle touching the x-axis and passing through the points (3,1) and (10,8).

Thanks.

Answer 1

(x-h)^2 + (y-k)^2 = r^2

Answer 2

yep
center; (h,k)

Answer 3

then what's next!?

Answer 4

then what's next!?

Answer 5

PLUG in (3,1)
(3-h)^2+(1-k)^2=r^2
plug in (10,8)
(10-h)^2+(8-k)^2=r^2
touching x-axis means we also have (x_0,0)
(x_0-h)^2+(0-k)^2=r^2

Answer 6

then what's next!?

Answer 7

last one of these was not so easy. i bet this one is not also

Answer 8

find the center: it will be equal distance from each point
use distance formula

Answer 9

try to figure out the radius given the points

Answer 10

i can't wait to see this. easy to say, not so easy to do

Answer 11

yea it is tricky! :p

Answer 12

You sure one of those isn't a center point? :(

Answer 13

so far i have gotten
\[h+k=11, k=\frac{9 \pm \sqrt{2r^2-49}}{2}, h=11-\frac{9 \pm \sqrt{2r^2-49}}{2}\]

Answer 14

i have an idea! (probably wrong)
put the center at (h,k) and point of tangency is (0,k) . distance from (9,k) to (h,k) is h

Answer 15

I can't see geometrically how it is possible to have both points in the circle and have the circle touch the - or + parts of the x-axis. In that case (3,1) would have to be part of a line segment that represents a chord in the circle. Now, you can make a circle that countains the two points and all you have to do is find the distance between the two points and then use the midpoint as your center (h,k) in the equation the tother member mentioned.

Answer 16

use the equations myininaya did above.
set them equal to each other, since r^2 = r^2
Then solve for h:
\[(3-h)^{2}+(1-k)^{2} = (10-h)^{2} +(8-k)^{2}\]
\[\rightarrow h = -k+11\]
Because its resting on x-axis, x_0 = h and r=k
\[h = 11-r\]
\[\rightarrow (3-(11-r))^{2} + (1-r)^{2} = r^{2}\]
solve for r
\[(r^{2}-16r+64)+(r^{2}-2r+1) =r^{2}\]
\[r^{2} -18r +65 = 0\]
\[(r-13)(r-5) = 0\]
\[r=5, h=6, k =5\]

Answer 17

remember teh distance is given by the formula

Answer 18

we have
\[h^2=(h-10)^2+(k-8)^2 = (h-3)^2+(k-1)^2\]
\[h^2=h^2-20h+100+k^2-16k+64\]
\[-20h +k^2-16k+164=0\]

Answer 19

oh i am too late. probably wouldn't have worked in any case!

Answer 20

Equation:
\[(x-6)^{2} + (y-5)^{2} = 25\]

Answer 21

especially since i had it tangent to the y - axis (doh)

Answer 22

satellite, yeah i think you had the right idea, just switched the axis

Answer 23

yeah no wonder i had a hard time drawing it!

Answer 24

no because its resting on x-axis Not y-axis

Answer 25

oh oops

Answer 26

i am going to try to redo this the way i did it before, see if i come up with an answer.

Answer 27

there were so many different circles that could have crossed those points!!

Answer 28

r = k is the kicker.

Answer 29

yeah i don't think you can do it unless you know its touching an axis

Answer 30

fantastic cow

Answer 31

yeah this is really nice