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If the circle x^2+y^2+2gx+2fy+c=0 bisect...

If the circle `x^2+y^2+2gx+2fy+c=0` bisects the circumference of the circle `x^2+y^2+2g^(prime)x+2f^(prime)y+c^(prime)=0` then prove that `2g^(prime)(g-g^(prime))+2f^(prime)(f-f^(prime))=c-c '`

Text Solution

Verified by Experts

Given circels are
`S_(2)-=x^(2)+y^(2)+2gx+2fy+c=0` (1)
`S_(2)-= x^(2)+y^(2)+2g'x+2f'y+c'=0` (2)
It is given that circle `S_(1)=0` bisects the circumference of the circle `S_(2)=0`. So, the common chord of circles paseses through the centre of the circle `S_(2)=0`.
Equation of common chord of the circles is `2x(g-g')+2y(f-f')+c-c'=0`.
This chord passes through the centre `( -g',-f')` of second circle.

`:. -2g'(g-g')-2f'(f-f')+c-c'=0`
`implies 2g'(g-g')+2f'(f-f')=c-c'`
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