QJEEAdv08Ma244f: JEE Advanced 2008 Mathematics - Class 11 JEEAdv08Ma244f Solution: Consider where p is a real number, and STATEMENT-1 : If line is …

Equation of circle C is

${(x + 3)^2} + {(y - 5)^2} = 9 + 25 - 30 = 4$

$\Rightarrow {(x + 3)^2} + {(y - 5)^2} = {2^2}$

Centre = (3, $-$ 5)

If L₁ is diameter, then it passes through the center (3, $-$ 5),

$\therefore$ $2(3) + 3( - 5) + p - 3 = 0 \Rightarrow p = 12$

$\therefore$ L₁ is $2x + 3y + 9 = 0$

and L₂ is $2x + 3y + 15 = 0$

Distance of centre of circle from {L_2}(d) = \left| {{{2(3) + 3( - 5) + 15} \over {\sqrt {{2^2} + {3^2}} }}} \right| = {6 \over {\sqrt {12} }} < 2 [radius of circle]

$\therefore$ L₂ is a chord of circle C.

Statement 2 is false.

Statement - 1 is true. If L₁ is a chord, L₂ may lie outside the circle.