A) 36
B) 72
C) 144
D) None of these
Correct Answer: B
Solution :
Given equation of line y = x is written in polar form is \[\frac{x}{\cos \theta }=\frac{y}{\sin \theta }=r\], where\[\theta =\frac{\pi }{4}\] For point\[P,\,\,r=6\sqrt{2}\]. Therefore, coordinates of\[P\]are given by \[\frac{x}{\cos \frac{\pi }{4}}=\frac{y}{\sin \frac{\pi }{4}}=6\sqrt{2}\] \[\Rightarrow \] \[x=6,\,\,y=6\] Since, point (6, 6) lies on \[{{x}^{2}}+{{y}^{2}}+2gx+2fy+c=0\] \[\therefore \] \[72+12(g+f)+c=0\] ... (i) Since,\[y=x\]touches the circle, therefore the equation\[2{{x}^{2}}+2x(g+f)+c=0\]has equal roots. \[\Rightarrow \] \[4{{(g+f)}^{2}}=8c\] \[\Rightarrow \] \[{{(g+f)}^{2}}=2c\] ... (ii) From Eq. (i) \[{{[12(g+f)]}^{2}}={{[-(c+72)]}^{2}}\] \[\Rightarrow \] \[144{{(g+f)}^{2}}={{(c+72)}^{2}}\] \[144(2c)={{(c+72)}^{2}}\] [from Eq.(u)] \[\Rightarrow \] \[{{(c-72)}^{2}}=0\] \[\Rightarrow \] \[c=72\]
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