A) \[-\,48\]
B) 80
C) \[-\,72\]
D) 54
E) 24
Correct Answer: A
Solution :
The equation of circle is \[3{{x}^{2}}+3{{y}^{2}}-6x-18y-7=0\] \[\Rightarrow \] \[{{x}^{2}}+{{y}^{2}}-2x-6y-\frac{7}{3}=0\] The centre of this circle is (1, 3). Also, two diameters of this circle are along the lines\[3x+y=q\]and\[x-3y={{c}_{2}}\] These two diameters should be passed from (1, 3). \[\therefore \] \[{{c}_{1}}=6\]and\[{{c}_{2}}=-8\] Hence, \[{{c}_{1}}{{c}_{2}}=6\times (-8)=-48\]
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