Question

Prove that the circles x²+y²+2ax+c² = 0 and x²+y²+2by+c² = 0 touches each other if 1/a² + 1/b² = 1/c². Please !!!

Answer 1

If the circles

x

2

+

y

2

+

2

a

x

+

c

2

=

0

and

x

2

+

y

2

+

2

b

y

+

c

2

=

0

touch externally prove that

1

a

2

+

1

b

2

=

1

c

4

Given that

the equation of first circle

x

2

+

y

2

+

2

a

x

+

c

2

=

0

=

(

x

+

a

)

2

+

y

2

+

c

2

−

a

2

=

0

=

(

x

+

a

)

2

+

y

2

=

(

√

a

2

−

c

2

)

2

So its center

C

1

→

(

−

a

,

0

)

and radius

r

1

=

√

a

2

−

c

2

and the equation of 2nd circle

x

2

+

y

2

+

2

b

y

+

c

2

=

0

=

x

2

+

(

y

+

b

)

2

+

c

2

−

b

2

=

0

=

x

2

+

(

y

+

b

)

2

=

(

√

b

2

−

c

2

)

2

So its center

C

2

→

(

0

,

−

b

)

and radius

r

2

=

√

b

2

−

c

2

As the two given circles touch externally, the distance between their centers

C

1

C

2

will be equal to sum of their radii.

C

1

C

2

=

r

1

+

r

2

⇒

√

a

2

+

b

2

=

√

a

2

−

c

2

+

√

b

2

−

c

2

Squaring both sides we get

a

2

+

b

2

=

a

2

−

c

2

+

b

2

−

c

2

+

2

√

a

2

−

c

2

√

b

2

−

c

2

⇒

2

c

2

=

2

√

a

2

−

c

2

√

b

2

−

c

2

⇒

c

4

=

(

a

2

−

c

2

)

(

b

2

−

c

2

)

⇒

c

4

=

a

2

b

2

−

b

2

c

2

−

a

2

c

2

+

c

4

⇒

b

2

c

2

+

a

2

c

2

=

a

2

b

2

⇒

1

a

2

+

1

b

2

=

1

c

2