If two circles x^2+y^2+2ax+c^2=0x2+y2+2ax+c2=0 and x^2+y^2+2by+c^2=0x2+y2+2by+c2=0 touch each other externally then prove that 1/(a^2)+1/(b^2)=1/(c^4)1a2+1b2=1c4 ?

2 Answers
Apr 27, 2018

Please refer to a Proof in Explanation.

Explanation:

Let the given circles be,

S_1 : x^2+y^2+2ax+c^2=0S1:x2+y2+2ax+c2=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!

Spread the Joy of Maths!

Apr 27, 2018

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 ?