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  • Equation of a circle that cuts the circle <img alt="\mathrm{x^2+y^2+2 g x+2 f y+c=0}" src="https://entrancecorner.oncodecogs.com/gif.latex?%5Cmathrm%7Bx%5E2&plus;y%5E

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

 

Option: 1

\mathrm{x^2+y^2+2 q x+2 f y+q^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

Clearly (-g, -f) will be the centre of required circle, 

Let the circle be \mathrm{(x+g)^2+(y+f)^2=d}

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

It has to cut the given circle orthogonally,

\mathrm{\begin{aligned} & \Rightarrow 2 g^2+2 f^2=c+g^2+f^2-d \\ & \Rightarrow d=c-g^2-f^2 \end{aligned}}

 

Posted by

Gaurav

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