A) \[{{x}^{2}}+{{y}^{2}}+2x-2y-62=0\]
B) \[{{x}^{2}}+{{y}^{2}}-2x+2y-62=0\]
C) \[{{x}^{2}}+{{y}^{2}}-2x+2y-47=0\]
D) \[{{x}^{2}}+{{y}^{2}}+2x-2y-47=0\]
Correct Answer: C
Solution :
The intersection point of diameters of a circle is the centre of a circle. The given equations of diameters are \[3x-4y-7=0\] ...(i) and \[2x-3y-5=0\] ...(ii) On solving Eqs. (i) and (ii). we get \[x=1\]and\[y=-1\] \[\therefore \]Centre of circle is (1, -1). Let r be the radius of circle, then \[\pi {{r}^{2}}=49\pi \] \[\Rightarrow \] \[r=7\]units \[\therefore \]Equation of required circle is \[{{(x-1)}^{2}}+{{(y+1)}^{2}}=49\] \[\Rightarrow \] \[{{x}^{2}}+{{y}^{2}}-2x+2y+1+1=49\] \[\Rightarrow \] \[{{x}^{2}}+{{y}^{2}}-2x+2y-47=0\]
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