Question

Find the equation of the circle that is tangent to the line whose equation is given by x + y = 2 and has its center at (3,5).

Answer 1

This is another version of finding the perpendicular to a line given a point and a line.

Then the circle will be x^2+y^2 = a^2 where a is the radius (distance)

Answer 2

real job here is to find the distance between the point and the line yes?

Answer 3

@estudier i think the equation will be
\[(x-3)^2+(y-5)^2=r^2\] we just need r

Answer 4

(x-3)+(y-5)=20

Answer 5

Yes, just a translation..

Answer 6

forgot to square the terms my bad

Answer 7

i am trying to think of a snap way to get the distance without dropping a perpendicular to the line from the point and then computing. how did you get
\[\sqrt{20}\]?

Answer 8

Midpoint of the line is the point where it would be tangential due to the slope being 1, so find the midpoint of the line (from x to y axis, I had (1,1), you should verify this, been up for 24 hours :S) and then use the distance formula from the center of the circle to the midpoint of the line. I may have gone wrong somewhere, but that was my methodology.

Answer 9

I agree though, your method would be better :). Find the perpendicular line first, then you can find a point on the line and the circle, follow up with the distance formula, and wholla your answer :).

Answer 10

i can do it as follows, but i thought there might be a snap way.
a) the slope of the line x+y=2 is -1 so the slope of the line giving the
distance from (3,5) to x +y=1 will be -1
2) the point on the line will be (x, 2-x)
3) so we find the piont via
\[\frac{5-(2-x)}{3-x}=1\]
\[\frac{3+x}{3-x}=1\]
\[x=0\] and so the point on the line closest to (3, 5) is (0,2)

Answer 11

but if i do this i don't get 20 for the square of the distance, get
18

Answer 12

U could translate the circle and line to (0,0) -> Pythagorus.

Answer 13

maybe i messed up

Answer 14

@estudier i think easier just to find square of distance because we already have center. all we need is r^2 to write equation

Answer 15

No, I agree with you, I guess I made a bad assumption, should've actually solved it the long way :).

Answer 16

oh ok. so answer is
\[(x-3)^2+(y-5)^2=18\]

Answer 17

@kris make any sense?

Answer 18

i got.
(x - h)2 + (y - k)2 = r2
(x - 3)2 + (y - 5)2 = (3sqrt(2))r2
(x - 3)2 + (y - 5)2 = 18

Answer 19

Pretty picture...

http://www.wolframalpha.com/input/?i=plot+x%2By%3D2%2C+%28x-3%29^2%2B%28y-5%29^2%3D+18

Answer 20

Just for the sake of it, take the circle at (0,0), x^2 +y^2 = r^2 and we want r.

Translate the line -> x+3+y+5 =2-> x + y =-6

After this, you have a right triangle at the origin bounded by the line with sides of 6 and r is the altitude so

r^2 = 6^2 - sqrt(72/2)^2 = 18 and r = sqrt 18