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If the circles x^2+y^2+2a x+c=0a n dx^2+...

If the circles `x^2+y^2+2a x+c=0a n dx^2+y^2+2b y+c=0` touch each other, then `1/(a^2)+1/(b^2)=1/c` (b) `1/(a^2)+1/(b^2)=1/(c^2)` (c) `a+b=2c` (d) `1/a+1/b=2/c`

A

`1//c`

B

c

C

`(1)/(c^(2))`

D

`c^(2)`

Text Solution

Verified by Experts

The correct Answer is:
A
`x^(2)+y^(2)+2ax+c=0,x^(2)+y^(2)+2by+c=0`
Common tangent at common point `ax-by=0`
Perpendicular distance from centre = radius
`x^(2)+y^(2)+2ax+c=0" "(-a,0)`
`|(-a^(2))/(sqrt(a^(2)+b^(2)))|=sqrt(a^(2)-c)impliesa^(4)=(a^(2)+b^(2))(a^(2)-c)`
`a^(4)=a^(4)+a^(2)b^(2)-c(a^(2)+b^(2))`
`(1)/( c )=(1)/(a^(2))+(1)/(b^(2))`
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Knowledge Check

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