1. A circles problem

If there are two circles such that $(x-a)^2 + (y-b)^2 &= c^2$ and $(x-b)^2 + (y-a)^2 &= c^2$ touch each other,

Then what can I write for " $a &=$? "

2. First You with simple algebraic steps write each equation in the form...

$\displaystyle \alpha (x,y,b)\ a^{2} + \beta (x,y,b)\ a + \chi (x,y,b) = 0$ (1)

... and then find a with the well known quadratic equation resolution formula...

Kind regards

$\chi$ $\sigma$

3. Originally Posted by PowerInside
If there are two circles such that $(x-a)^2 + (y-b)^2 &= c^2$ and $(x-b)^2 + (y-a)^2 &= c^2$ touch each other,

Then what can I write for " $a &=$? "
The two circles have the following properties:

- the circles are congruent
- common tangent of both circles is the line y = x
- each circle is the image of the other circle by reflection over the line y = x
- the distance between the centers equals the diameter of one circle

$d = \sqrt{(a-b)^2+(b-a)^2}=2c$

Therefore:

$\sqrt{2(a-b)^2}=2c~\implies~2(a-b)^2=4c^2$

Solve for a: $a = b+c\sqrt{2}$

4. Shouldnt the final answer be $a &= b \pm c\sqrt{2}$

5. Originally Posted by chisigma
First You with simple algebraic steps write each equation in the form...

$\displaystyle \alpha (x,y,b)\ a^{2} + \beta (x,y,b)\ a + \chi (x,y,b) = 0$ (1)

... and then find a with the well known quadratic equation resolution formula...

Kind regards

$\chi$ $\sigma$
I don't understand this approach. Is it relevent for solving this problem? I did not understand the 3 variable equation you gave to use? ??

Sorry for being a newbie , But I have difficulty in understanding this one. And im trying to sharpen my math skills..

6. Originally Posted by PowerInside
Shouldnt the final answer be $a &= b \pm c\sqrt{2}$
You are right. Value of a will be different depending on the circles touching internally or externally.

7. Originally Posted by sa-ri-ga-ma
You are right. Value of a will be different depending on the circles touching internally or externally.
I don't want to pick at you, but:
If there are two circles such that $(x-a)^2 + (y-b)^2 &= c^2$ and $(x-b)^2 + (y-a)^2 &= c^2$ touch each other,

Then what can I write for " a = ? "
According to the text of the question both circles have the same radius. I tried to find two congruent circles where one circle touches the other internally. To be honest I wasn't very successful.

To be exact you have to consider 3 cases:

1. a > b, that means the circle with M(a, b) is placed below the line y = x

2. a = b, that means both circles have the same midpoint on the line y = x (and only in this case there could be an internal touching)

3. a < b, that means the circle with M(a, b) is placed above the line y = x

I've used only the first case because the 3rd case is the reflection of the 1st case over the line y = x.

8. Originally Posted by earboth
I don't want to pick at you, but:

According to the text of the question both circles have the same radius. I tried to find two congruent circles where one circle touches the other internally. To be honest I wasn't very successful.
.
Yes. You are right.

9. As earboth shows a= b +crad2 when a is greater than b, a=b -crad 2 when b is greater than a . Another point y=x and y=-xand the two tangent circles can be in any of the 4 quadrants.

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