question on circles...

Hello just another question on circles:

Verify that the circle with equation x² + y² - 2rx - 2ry + r² = 0 touches both the coordinate axes. Find the radii of the two circles which pass through the point (16,2) and touch both the coordinate axes.

Thanks
From John

Reply 1

m00c0w06

Don't know if this is an accurate proof or not but here is what I got.

Complete the square on x² + y² - 2rx - 2ry + r² = 0 to get (x-r)^2 + (y-r)^2 = r^2

So the center of the circle is at (r,r) and the radius is sqt(r^2) = r. So if the center is at (r,r) and the radius is r then you can see that it will touch both coordinate axes.

Reply 2

Neo1

I dont think that's right i think the answer is 10 and 26 something like that but I dont know how to get it....

can anyone help please

Thanks
From John

Reply 3

m00c0w06

Yes the answer to the second part is 26 and 10.

We know that the center is at (r,r), the circle goes through the point (16,2) and the radius is r. So to work out the length of the radius we need to find the length of the line segment joining (r,r) and (16,2)

Using Pythagoras we get...

r^2 = (2-r)^2 + (16-r)^2
r^2 = 4 - 4r + r^2 + 256 - 32r + r^2
r^2 - 36r + 260 = 0
(r-26)(r-10) = 0
r = 26 and r = 10

Reply 4

Neo1

Another question which I find tricky is this one:

Circle C1 has equation x² + y² + 4x - 6y - 12 = 0 and circle C2 has equation x² + y² - 20x + 12y + 100 = 0. Point P lies on C1 and point Q lies on C2. The distance between P and Q is denoted by d. Show that 4 < d < 26

Well I think I should start finding the centre of the circle and radius for each but then after that what?

Thanks a lot
From John

Reply 5

m00c0w06

Assuming the two circles don't intersect (I don't know whether they do or not at this point)...

Once you've found the centers are the radii then the smallest value of d is the length of the line segment between the radius of circle C1 and the radius of circle C2 minus (the radius of C1 + the radius of C2). The largest value of d is the smallest value of d + radius of C1 + radius of C2. Are you sure the inequality isn't actually 4 <= d <= 26?

Reply 6

m00c0w06

For Circle C1...

x^2 + 4x + y^2 - 6y = 12
(x+2)^2 + (y-3)^2 = 25

Centre is (-2,3)
Radius is 5

For Circle C2...

x^2 + y^2 - 20x + 12y = -100
(x-10)^2 + (y+6)^2 = 36

Centre is (10,-6)
Radius is 6

Let A be the center of C1 and B be the center of C2.

(AB)^2 = (-6-3)^2 + (10 - -2)^2
(AB)^2 = 81 + 144 = 225
AB = 15

This is the distance between the two radii therefore if we subtract the sum of the two radii we will get the minimum distance between the circumferences of the two circles...

15 - (6+5) = 4

If we now add the two radii we'll get the maximum distance of the two points

15 + (6+5) = 26

Therefore

4 <= d <= 26