Consider $L_1: 2 x+3 y+p-3=0 ; L_2: 2 x+3 y+p+3=0$ where $p$ is a real number and $C: x^2+y^2+6 x-10 y+30=0$…
Consider $L_1: 2 x+3 y+p-3=0 ; L_2: 2 x+3 y+p+3=0$ where $p$ is a real number and $C: x^2+y^2+6 x-10 y+30=0$. Statement 1 If line $L_1$ is a chord of circle $C$, then line $L_2$ is not always a diameter of circle $C$. Statement 2 If line $L_1$ is a diameter of circle $C$, then line $L_2$ is not a chord of circle $C$.
Statement 1 is true, Statement 2 is true, Statement 2 is a correct explanation for Statement 1.
Statement 1 is true, Statement 2 is true, Statement 2 is not a correct explanation for Statement 1.
Statement 1 is true, Statement 2 is false.
Statement 1 is false, Statement 2 is true
Solution
Equation of circle $C$ is
$
\begin{aligned}
(x+3)^2+(y-5)^2 & =9+25-30=4 \\
\Rightarrow \quad \quad(x+3)^2+(y-5)^2 & =2^2 \\
\text { Centre } & =(3,-5)
\end{aligned}
$
If $L_1$ is diameter, then $2(3)+3(-5)+p-3=0 \Rightarrow p=12$
$\therefore \quad L_1$ is $2 x+3 y+9=0$
and
$L_2$ is $2 x+3 y+15=0$
Distance of centre of circle from $L_2=\left|\frac{2(3)+3(-5)+15}{\sqrt{2^2+3^2}}\right|=\frac{6}{\sqrt{13}} < 2 \quad$ [radius of circle]
$\therefore L_2$ is a chord of circle $C$.
Statement 2 is false