Question

Two diameters of the circle $3x^2+3y^2-6x$ $-18y-7=0$ are along the lines $3x+y=q$ and $x-3y=c_2$ .Then the value of $c_1{c}_2$ is

• A. $-48$
• B. 80
• C. $-72$
• D. 54
• E. 24

Answer 1

Answer: (A)

The equation of circle is $3x^2+3y^2-6x-18y-7=0$
$\Rightarrow$ $x^2+y^2-2x-6y-\frac{7}{3}=0$
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, $c_1{c}_2=6\times (-8)=-48$