If there are two circles such that $\displaystyle (x-a)^2 + (y-b)^2 &= c^2$ and $\displaystyle (x-b)^2 + (y-a)^2 &= c^2$ touch each other,
Then what can I write for " $\displaystyle a &= $? "
If there are two circles such that $\displaystyle (x-a)^2 + (y-b)^2 &= c^2$ and $\displaystyle (x-b)^2 + (y-a)^2 &= c^2$ touch each other,
Then what can I write for " $\displaystyle a &= $? "
First You with simple algebraic steps write each equation in the form...
$\displaystyle \displaystyle \alpha (x,y,b)\ a^{2} + \beta (x,y,b)\ a + \chi (x,y,b) = 0$ (1)
... and then find a with the well known quadratic equation resolution formula...
Kind regards
$\displaystyle \chi$ $\displaystyle \sigma$
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
$\displaystyle d = \sqrt{(a-b)^2+(b-a)^2}=2c$
Therefore:
$\displaystyle \sqrt{2(a-b)^2}=2c~\implies~2(a-b)^2=4c^2$
Solve for a: $\displaystyle a = b+c\sqrt{2}$
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.If there are two circles such that $\displaystyle (x-a)^2 + (y-b)^2 &= c^2$ and $\displaystyle (x-b)^2 + (y-a)^2 &= c^2$ touch each other,
Then what can I write for " a = ? "
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.