A) \[{{x}^{2}}+{{y}^{2}}+2x-2y-23=0\]
B) \[{{x}^{2}}+{{y}^{2}}-2x-2y-23=0\]
C) \[{{x}^{2}}+{{y}^{2}}+2x-2y-23=0\]
D) \[{{x}^{2}}+{{y}^{2}}-2x+2y-23=0\]
Correct Answer: D
Solution :
According to question two diameters of the circle are \[2x+3y+1=0\] and \[3x-y-4=0\] Solving, we get \[x=1,\,y=-1\] \[\therefore \] Centre of the circle is (1, ? 1) Given \[2\pi r=10\pi \Rightarrow r=5\] \[\therefore \] Required circle is \[{{(x-1)}^{2}}+{{(y+1)}^{2}}={{5}^{2}}\] or \[{{x}^{2}}+{{y}^{2}}-2x+2y-23=0\].
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