A) \[{{x}^{2}}+{{y}^{2}}+6x-16y+28=0\]
B) \[{{x}^{2}}+{{y}^{2}}-6x-16y+28=0\]
C) \[{{x}^{2}}+{{y}^{2}}+6x+6y+28=0\]
D) \[{{x}^{2}}+{{y}^{2}}-6x-6y-28=0\]
Correct Answer: A
Solution :
Equation of line which is perpendicular to the tangent\[2x-y-1=0\]is\[x+2y+k=0\]. Equation of tangent passes through the point (3,5). \[\therefore \] \[3+2\times 5+k=0\] \[\Rightarrow \] \[k=-13\] \[\therefore \]Centre of circle will lie on the perpendicular line\[x+2y=13\]. Also, centre of circle lies on the line\[x+y=5\]. So, on solving these two equations, we get \[x=-3,y=8\] Centre\[=(-3,8)\] and radius\[=\sqrt{{{(3+3)}^{2}}+{{(5-8)}^{2}}}\] \[=\sqrt{{{6}^{2}}+{{3}^{2}}}\] \[=\sqrt{45}\] \[\therefore \]Equation of circle is \[{{(x+3)}^{2}}+{{(y-8)}^{2}}={{(\sqrt{45})}^{2}}\] \[\Rightarrow \] \[{{x}^{2}}+9+6x+{{y}^{2}}+64-16y=45\] \[\Rightarrow \] \[{{x}^{2}}+{{y}^{2}}+6x-16y+28=0\]
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