Question

If two circles `x^2+y^2+2ax+c^2=0` and `x^2+y^2+2by+c^2=0` touch each other externally then prove that `1/(a^2)+1/(b^2)=1/(c^4)` ?

Answer 1

Please refer to a Proof in Explanation.

Explanation:

Let the given circles be,

`S_1 : x^2+y^2+2ax+c^2=0`.

`:. (x+a)^2+y^2=(sqrt(a^2-c^2))^2, (a^2-c^2) gt 0`.

Clearly, the centre `C_1` and the radius `r_1` of `S_1` are,

`C_1=C_1(-a,0), and, r_1=sqrt(a^2-c^2)`.

Analogously for `S_2 : x^2+y^2+2by+c^2=0`, we have,

`C_2=C_2(0,-b), and, r_2=sqrt(b^2-c^2), (b^2-c^2) gt 0`.

Given that `S_1 and S_2` touch each other externally.

Evidently, `"distance "C_1C_2=r_1+r_2`.

Utilising `C_1, C_2, r_1, r_2`, we get, from the above eqn.,

`(0-(-a))^2+(-b-0)^2=(r_1+r_2)^2, or,`

`r_1^2+2r_1r_2+r_2^2=a^2+b^2`.

`:. (a^2-c^2)+2r_1r_2+(b^2-c^2)=a^2+b^2`.

`:. r_1r_2=c^2`.

`:. r_1^2r_2^2=c^4`.

`:. (a^2-c^2)(b^2-c^2)=c^4`.

`:. a^2b^2-a^2c^2-b^2c^2+c^4=c^4`.

`:. a^2b^2=a^2c^2+b^2c^2=(a^2+b^2)c^2`.

Dividing throughout by `a^2b^2c^2!=0" (Why?)"`, we conclude,

`1/c^2=(a^2+b^2)/(a^2b^2)=1/a^2+1/b^2`, as desired!

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Answer 2

Please see the explanation below.

Explanation:

The first circle is

`x^2+2ax+y^2=-c^2`

`x^2+2ax+a^2+y^2=a^2-c^2`

`(x-a)^2+y^2=a^2-c^2`

The center is `(a,0)` and the radius is `sqrt(a^2-c^2)`

The second circle is

`x^2+y^2+2by=-c^2`

`x^2+y^2+2by+b^2=b^2-c^2`

`x^2+(y+b^2=b^2-c^2`

The center is `(0,b)` and the radius is `sqrt(b^2-c^2)`

If the circles touch each other externally, then

`"distance between the centers "=" sum of the radii"`

`sqrt(a^2+b^2)=sqrt(a^2-c^2)+sqrt(b^2-c^2)`

Squaring both sides,

`(sqrt(a^2+b^2))^2=(sqrt(a^2-c^2)+sqrt(b^2-c^2))^2`

`a^2+b^2=a^2-c^2+b^2-c^2-2sqrt((a^2-c^2)(b^2-c^2))`

`c^2=-sqrt((a^2-c^2)(b^2-c^2))`

Squaring both sides

`c^4=a^2b^2-a^2c^2-b^2c^2+c^4`

`a^2b^2=a^2c^2+b^2c^2`

Dividing by `a^2b^2c^2`

`1/c^2=1/a^2+1/b^2`

Are you sure of the `1/c^4` ?