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Equation of a circle that cuts the circle \mathrm{x^{2}+y^{2}+2 g x+2 f y+c=0}, lines \mathrm{x=-g} and \mathrm{y=-f} orthogonally, is;

Option: 1

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


Option: 2

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


Option: 3

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


Option: 4

 none of these


Answers (1)

best_answer

 Lines \mathrm{x=-g} and \mathrm{y=-f} will be diameter for the circle. Equation of circle will be

\mathrm{(x+g)^{2}+(y+f)^{2}=r^{2}}. Now this intersect \mathrm{x^{2}+y^{2}+2 g x+2 f y+c=0}  orthogonally

Hence 2g.g. \mathrm{+2 f . f=c+k}

\mathrm{\Rightarrow \quad k=2 g^{2}+2 f^{2}-c}

Hence (D) is the correct answer.

 

Posted by

shivangi.shekhar

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