# Geometry problem

• May 18th 2006, 09:14 AM
UFOKatarn
Geometry problem
First, hello all. Great I found this forum.

-----------------------------------------------------------

we have two vertices:

A(1,1)
B(0,2)

and there is also a circle K:

K: (x-5)^2 + (x-5)^2 = 16

Now I have to find a circle G: (x-a)^2 + (x-b)^2 = r^2

This circle (G) has to:
- have both vertices (A and B) on its edge
- "touch" the circle K in only one vertex

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G1: (x-1)^2 + (x-2)^2 = 1

G2: (x-4)^2 + (x-5)^2 = 25

-------------------------------------------------------------------

I have the answers all right, however I don't know how to calculate them.

And excuse my english, because I am not very familiar with english mathematical terminology.

Thank you,
UFOKatarn
• May 18th 2006, 04:21 PM
ticbol
Quote:

Originally Posted by UFOKatarn
First, hello all. Great I found this forum.

-----------------------------------------------------------

we have two vertices:

A(1,1)
B(0,2)

and there is also a circle K:

K: (x-5)^2 + (x-5)^2 = 16

Now I have to find a circle G: (x-a)^2 + (x-b)^2 = r^2

This circle (G) has to:
- have both vertices (A and B) on its edge
- "touch" the circle K in only one vertex

-------------------------------------------------------------------

G1: (x-1)^2 + (x-2)^2 = 1

G2: (x-4)^2 + (x-5)^2 = 25

-------------------------------------------------------------------

I have the answers all right, however I don't know how to calculate them.

And excuse my english, because I am not very familiar with english mathematical terminology.

Thank you,
UFOKatarn

Okay, let me change then some of your data.
--vertex == point.
--edge == circumference.
--Circle K should be: (x-5)^2 +(y-5)^2 = 16 ----(1)
--Circle G: (x-a)^2 +(y-b)^2 = r^2 ----------(2)
--"touch only in one vertex" == "is tangent to"
--mistery == problem only, :-)

If circle G passes through the points (1,1) and (0,2), then,
(1-a)^2 +(1-b)^2 = r^2 ----at (1,1)------(i)
(0-a)^2 +(2-b)^2 = r^2 ----at (0,2)------(ii)
Eliminate r, (i) minus (ii),
(1-a)^2 -(0-a)^2 +(1-b)^2 -(2-b)^2 = 0
(1-a)^2 -(-a)^2 = (2-b)^2 -(1-b)^2
1 -2a +a^2 -a^2 = 4 -4b +b^2 -(1 -2b +b^2)
1 -2a = 3 -2b
1 -2a -3 = -2b
-2a -2 = -2b
b = a+1`-----------**
That means the center of circle G is at point (a,b) == (a,a+1) --------**
Meaning, circle G is
(x -a)^2 +(y -(a+1))^2 = r^2
(x-a)^2 +(y-a-1)^2 = r^2 ---------(2.1)
We still have two unknowns: a and r.

Let us express r in terms of "a".
At point (1,1), using Eq.(2.1),
(1-a)^2 +(1-a-1)^2 = r^2
(1 -2a +a^2) +(a^2) = r^2
2a^2 -2a +1 = r^2 ----------------**

Now, if circle G is tangent to circle K, then the distance between their centers is minimum at the point of tangency. This minimum distance is a straight line joining the centers of the two circles. It is the sum of the radius of K and the radius of G. Thus,
--radius of circle K = sqrt(16) = 4.
--radius of circle G = sqrt(r^2) = sqrt(2a^2 -2a +1).
--center of circle K = (5,5).
--center of cicle G = (a,a+1).
--distance between the two centers = distance between points (5,5) and (a,a+1).
So,
4+r = sqrt[(5-a)^2 +(5 -(a+1))^2]
4+r = sqrt[(5-a)^2 +(5-a-1)^2]
4+r = sqrt[(25 -10a +a^2) +(16 -8a +a^2)]
4+r = sqrt[41 -18a +2a^2]
Plugging in the value of r in terms of "a",
4 +sqrt[2a^2 -2a +1] = sqrt[41 -18a +2a^2]
Rationalize one of the radicals, square both sides,
16 +8sqrt[2a^2 -2a +1] +(2a^2 -2a +1) = 41 -18a +2a^2
8sqrt[2a^2 -2a +1] = 41 -18a +2a^2 -16 -2a^2 +2a -1
8sqrt[2a^2 -2a +1] = 24 -16a
Divide both sides by 8,
sqrt[2a^2 -2a +1] = 3 -2a
Square both sides,
2a^2 -2a +1 = 9 -12a +4a^2
Bring them all to the lefthand side,
2a^2 -2a +1 -9 +12a -4a^2 = 0
-2a^2 +10a -8 = 0
Divide both sides by -2,
a^2 -5a +4 = 0
(a-4)(a-1) = 0
a = 4 or 1 -----------***

When a=4,
b = a+1 = 5
Hence, center of circle G is (a,b) == (4,5)
r = sqrt[2a^2 -2a +1] = sqrt[2(4^2) -2(4) +1] = 5
Therefore, circle G is
(x-4)^2 +(y-5)^2 = 5^2

When a=1,
b = a+1 = 2
Hence, center of circle G is (a,b) == (1,2)
r = sqrt[2a^2 -2a +1] = sqrt[2(1^2) -2(1) +1] = 1
Therefore, circle G is
(x-1)^2 +(y-2)^2 = 1^2

------------
I did the (5 -a -1)^2 on paper by "long" multiplication.
• May 18th 2006, 06:43 PM
ThePerfectHacker
I solve this problem more elegantly I am going to state a test to determine whether two circles "touch", i.e. tangent to each other.

Let us assume you have two circles with $R$ and $r$ with $R>r$.
There are two possibilities with tangent circles.
Internally tangent-meaning the smaller is inside the larger and touches it.
Externally tangent-meaning the smaller and larger are outside and touch at only one point.
In the first case the distance between the radii is,
$R-r$.
In the second case the distance between the radii is,
$R+r$.
As a result we have a test to determine whether two circles are tangent. You determine whether or not the distance between the two radii is $R\pm r$.
--------
Returning to the problem.
You have a circle,
$(x-5)^2+(y-5)^2=4^2$ (1)
You need to find a circle(s),
$(x-a)^2+(y-b)^2=r^2$ (2)
Which satisfies 2 conditions:
1)Is tangent to (1)
2)Passes through $(1,1),(0,2)$

For condition 1) to be true we need that (1) is tangent to (2), using the test I said in the introduction we need that the distance between the radii is $|r\pm 4|$. The reason why the absolute value is because we do not know whether $r \mbox{ or }4$ is larger thus we need the positive value of the difference, i.e. absolute value.
Now, the center of (1) is at (5,5) and the distance of (2) is at (a,b). Thus by the distance formula,
$\sqrt{(a-5)^2+(b-5)^2}=|r\pm 4|$
Square both sides, note that absolute value disappears,
$(a-5)^2+(b-5)^2=(r\pm 4)^2$

For condition 2) point (1,1) and (0,2) are on (2) thus,
$(1-a)^2+(1-b)^2=r^2$
$(0-a)^2+(2-b)^2=r^2$
-------
As a result we have a triple equation with 3 unknowns,
$\left\{ \begin{array}{c}
(a-5)^2+(b-5)^2=(r\pm 4)^2 \\
(a-1)^2+(b-1)^2=r^2 \\
a^2+(b-2)^2=r^2
$

Solving this should give you the results.
Note, you need to consider two cases for $\pm$ one for plus and other for minus.