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  • Equation of the circle which passes through the origin and cuts orthogonally each of the circle <img alt="\mathrm{ x^2+y^2-8 y+12=0}" src="https://entrancecorner.oncodecogs.com/gif.latex?%5Cma

Equation of the circle which passes through the origin and cuts orthogonally each of the circle \mathrm{ x^2+y^2-8 y+12=0} and \mathrm{x^2+y^2-4x-6 y-3=0 }is
 

Option: 1

\mathrm{ x^2+y^2+6 x+3 y=0}
 


Option: 2

\mathrm{ x^2+y^2+3 x-6 y=0}


Option: 3

\mathrm{ x^2+y^2+6 x-3 y=0}


Option: 4

\mathrm{x^2+y^2-3 x+6 y=0}


Answers (1)

best_answer

Equation of the circle through the origin can be written as
\mathrm{x^2+y^2+2 g x+2 f y=0 }

It cuts the two given circles orthogonally

so    \mathrm{2.0. g-2.4 .f=12+0 }

and \mathrm{-4 g-6 f=-3+0 }

Solving \mathrm{g=3, f=\frac{-3}{2}}

\therefore \quad The equation of circle is \mathrm{ x^2+y^2+6 x-3 y=0}

Posted by

Ritika Jonwal

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