Question

Let $f ( 0 , \lambda ) = 0$ has equal roots $\lambda = 2,2$ and $f ( \lambda , 0 ) = 0$ has roots then the centre of the circle is

• A. $\left( 2 , \frac { 29 } { 10 } \right)$
• B. $\left( \frac { 29 } { 10 } , 2 \right)$
• C. $\left( - 2 , \frac { 29 } { 10 } \right)$
• D. None

Answer 1

Answer: (B)

$\left( \frac { 29 } { 10 } , 2 \right)$

Answer 2

$\because f ( x , y ) \equiv x ^ { 2 } + y ^ { 2 } + 2 g x + 2 f y + c = 0$Now, $f ( 0 , \lambda ) \equiv \lambda ^ { 2 } + 2 f \lambda + c = 0$ and its roots are 2, 2.

$\therefore 2 + 2 = - 2 f , 2 \times 2 = c$ i.e.

$f ( \lambda , 0 ) \equiv \lambda ^ { 2 } + 2 g \lambda + c = 0$ and its roots are $\frac { 4 } { 5 } , 5$.

$g = \frac { - 29 } { 10 } , c = 4$.

Hence, centre of the circle $= ( - g , - f ) = \left( \frac { 29 } { 10 } , 2 \right)$