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Q.
Two diameters of the circle $ 3{{x}^{2}}+3{{y}^{2}}-6x $ $ -18y-7=0 $ are along the lines $ 3x+y=q $ and $ x-3y={{c}_{2}} $ .Then the value of $ {{c}_{1}}{{c}_{2}} $ is
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 $