A) \[{{y}^{2}}=16x+4\]
B) \[{{x}^{2}}=16y\]
C) \[{{x}^{2}}=16y+4\]
D) \[{{y}^{2}}=16x\]
Correct Answer: D
Solution :
Let the circle be \[{{x}^{2}}+{{y}^{2}}+2gx+2fy+c=0\] .....(i) It cuts the circle \[{{x}^{2}}+{{y}^{2}}-20x+4=0\] orthogonally \[\therefore \] \[2(-10g+0\times f)=c+4\] \[\Rightarrow \] \[-20g\,=\,c+4\] .....(ii) Circle (i) touches the line \[x=2\], \\[x+0y-2=0\] \[\therefore \] \[\left| \frac{-g+0-2}{\sqrt{{{1}^{2}}+{{0}^{2}}}} \right|=\sqrt{{{g}^{2}}+{{f}^{2}}-c}\] \[\Rightarrow \] \[{{(g+2)}^{2}}={{g}^{2}}+{{f}^{2}}-c\]\[\Rightarrow \] \[4g+4={{f}^{2}}-c\] .....(iii) Eliminating c from (ii) and (iii), we get \[-16g-4={{f}^{2}}-4\] \[\Rightarrow \] \[{{f}^{2}}+16g=0\] Hence the locus of \[(-g,\,-f)\] is \[{{y}^{2}}-16x=0\].
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