Question

Find the equation of the circle of radius sqr(10) that are tangent to 3x+y=6 at (3,-3)

Answer 1

Okay, so we have a point on the surface of the circle, right? That's (3,-3). And we have its radius.
The equation of this circle is (x-h)^2+(y-k)^2=10, where (h,k) is the center and \[\sqrt{10}\] is the radius.

Answer 2

So, what we really need to find is the center.

Answer 3

working it out, I was able to get k=1/3h-4

Answer 4

then plug that into the original eq of (3-h)^2+(-3-k)^2=10

Answer 5

Since we know that the line represented by 3x+y=6 is tangent to the circle, then the line perpendicular to 3x+y=6 at (3,-3) will pass right through the center of the circle. We also know that it will do so at a point that is \[\sqrt{10}\] distant from (3,-3).

Answer 6

Donutlicious, that's only going to get you something like 1=1 or some such.

Answer 7

So, to get a line perpendicular to 3x+y=6, let's put it into point-slope format:
y=-3x-6

Alright, so a line perpendicular to that one would be y = 3x, since it would have the opposite slope. However, we want the line to intersect at the point (3,-3), so we'll need to add some constant b and solve for it:
y = 3x + b
-3 = 3(3) + b
-3 - 9 = b
b = -12

Answer 8

@jonnybot the perpendicular line would have slope
\[\frac{1}{3}\] but i think you are on the right track

Answer 9

Satellite can u come help me when u r finished? ;D

Answer 10

Okay, so we've got out perpendicular line! Now we know that the center of the circle is sqrt(10) away from (3,-3) and that it lies on the line y=3x-12. Using the distance formula between the center of the circle and the line, we know that:
sqrt(10) = sqrt{(3-h)^2 + (-3-k)^2}
Where (h,k) is the center of the circle).

Answer 11

Ah I see. I'm jotting this all down as we go through this. Thank you for your time by the way :)

Answer 12

Also, I'm pretty sure satellite is wrong, and that the perpendicular line would have slope of 3.

Answer 13

hey don't put down satellite!

Answer 14

pretty sure the perpendicular line has slope
\[\frac{1}{3}\] and equation
\[y+3=\frac{1}{3}(x-3)\] or
\[y=\frac{1}{3}x-4\] so any point on the line will look like
\[(x. \frac{1}{3}x-4)\] and if it is
\[\sqrt{10}\] away from
\[(3,-3)\] then you know that
\[(x-3)^2+\frac{1}{3}(x-1)^2=10\] solve to get
\[x=0, x=\frac{24}{5}\]

Answer 15

@jonnybot line perpendicular to a line with slope m has slope
\[-\frac{1}{m}\] and so if your line has slope -3, perpendicular line will have slope
\[\frac{1}{3}\]

Answer 16

And Satellite... look at http://www.wolframalpha.com/input/?i=graph+y%3D-3x+and+y%3D3x . Tell me those lines aren't perpendicular. Then look at http://www.wolframalpha.com/input/?i=graph+y%3D-3x+and+y%3D%281%2F3%29x and try to tell me that they are.

Answer 17

By the way, I'm not trying to talk smack here or anything.

Answer 18

Wait... satellite is right...
http://www.wolframalpha.com/input/?i=graph+y%3D-3x+and+y%3D%281%2F3%29x+from+x%3D-1+to+x+%3D1+and+y+%3D+-1+to+y+%3D+1

I was deluded. :)

Answer 19

Ah man, he left :( I still needed help.