Prove that the length of the common chord of the two circles :
`(x-a)^(2)+(y-b)^(2)=c^(2)and(x-b)^(2)+(y-a)^(2)=c^(2)" is "sqrt(4c^(2)-2(a-b)^(2))`
Find also the condition when the given circles touch.

Options

A. `2sqrt(c^(2)-(a-b)^(2))`
B. `sqrt(4c^(2)-2(a-b)^(2))`
C. `sqrt(2c^(2)-(a-b)^(2))`
D. `sqrt(4c^(2)+2(a-b)^(2))`

Answer

B

Explanation

The equation of the common chord PQ of the circles
`(x-a)^(2)+(y-b)^(2)=c^(2) and (x-b)^(2)+(y-a)^(2)=c^(2)` is
x-y=0 [Using: `S_(1)-S_(2)=0`]

Now,
` C_(1)M`=Length of perpendicular from (A, b) to x-y=0
`rArr C_(1)M=|(a-b)/(sqrt(2))|`
`:. PQ=2PM=2sqrt(C_(1)P^(2)-C_(1)M^(2))=2sqrt(c^(2)-((a-b)^(2))/(2))`
`rArr PQ=2 sqrt((2c^(2)-(a-b)^(2))/(2))=sqrt(4c^(2)-2(a-b)^(2))`