JEE MATHEMATICS MCQ: CIRCLES

1 Easy

1 of 1

Application +4 | -1

The circle

x^2+y^2+2 g x+2 f y+c=0

bisects the circumference of the circle

x^2+y^2+2 g^{\prime} x+2 f^{\prime} y+c^{\prime}=0

, if

View answer and explanation

Correct option: A

2 g^{\prime}\left(g-g^{\prime}\right)+2 f^{\prime}\left(f-f^{\prime}\right)=c-c^{\prime}

Option A (

2 g^{\prime}\left(g-g^{\prime}\right)+2 f^{\prime}\left(f-f^{\prime}\right)=c-c^{\prime}

) is correct because it represents the condition for the first circle to bisect the circumference of the second circle.

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9 \pi