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  • The locus of the centre of a circle which cuts orthogonally the circle <img alt="\mathrm{x^2+y^2-20 x+4=0}" src="https://entrancecorner.oncodecogs.com/gif.latex?%5Cmathrm%7Bx%5E2&plus;y%5E2-20%

The locus of the centre of a circle which cuts orthogonally the circle \mathrm{x^2+y^2-20 x+4=0} and which touches \mathrm{x=2} is

Option: 1

\mathrm{y^2=16 x+4}


Option: 2

\mathrm{x^2=16 y}


Option: 3

\mathrm{x^2=16 y+4}


Option: 4

\mathrm{y^2=16 x}


Answers (1)

Let the circle be \mathrm{x^2+y^2+2 g x+2 f y+c=0}                          \mathrm{...(i)}

It cuts the circle \mathrm{x^2+y^2-20 x+4=0} orthogonally

\mathrm{ \therefore 2(-10 g+0 \times f)=c+4 \Rightarrow-20 g=c+4 }                    \mathrm{...(ii)}

Circle (i) touches the line \mathrm{x=2 ; \quad \therefore \quad x+0 y-2=0}

\mathrm{\therefore\left|\frac{-g+0-2}{\sqrt{1}}\right|=\sqrt{g^2+f^2-c} \Rightarrow(g+2)^2=g^2+f^2-c \Rightarrow 4 g+4=f^2-c}                   \mathrm{...(iii)}

Eliminating c from (ii) and (iii), we get \mathrm{-16 g+4=f^2+4 \Rightarrow f^2+16 g=0}

Hence the locus of \mathrm{(-g,-f) \text { is } y^2-16 x=0 \Rightarrow y^2=16 x}

Posted by

Ramraj Saini

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