1. Write the circle in standard form

Given $x^2+y^2+6x-10y+30=0$ Complete squares: $x^2+6x+y^2-10y+30=0$ $(x+3)^2-9+(y-5)^2-25+30=0$ $(x+3)^2+(y-5)^2=4$

So the circle has:

• Centre $(-3,5)$
• Radius $r=2$

2. Observe the two lines

$L_1: 2x+3y+p-3=0$ $L_2: 2x+3y+p+3=0$

These are parallel lines.

For a line $2x+3y+c=0$, distance from the centre $(-3,5)$ is $d=\frac{|2(-3)+3(5)+c|}{\sqrt{2^2+3^2}}=\frac{|-6+15+c|}{\sqrt{13}}=\frac{|9+c|}{\sqrt{13}}$

3. Condition for $L_1$ to be a chord

For $L_1$, we have $c=p-3$. Hence distance from centre: $d_1=\frac{|9+(p-3)|}{\sqrt{13}}=\frac{|p+6|}{\sqrt{13}}$

A line is a chord of the circle if it intersects the circle in two distinct points, i.e. $d_1<r=2$ So, $\frac{|p+6|}{\sqrt{13}}<2$ $|p+6|<2\sqrt{13}$

Now for $L_2$, $c=p+3$, so distance from centre is $d_2=\frac{|9+(p+3)|}{\sqrt{13}}=\frac{|p+12|}{\sqrt{13}}$

A line can be a diameter only if it passes through the centre, i.e. $d_2=0$ which gives $p+12=0 \Rightarrow p=-12$

But if $p=-12$, then for $L_1$, $d_1=\frac{|-12+6|}{\sqrt{13}}=\frac{6}{\sqrt{13}}<2$ (since $36<52$), so indeed $L_1$ is a chord.

Thus when $L_1$ is a chord, $L_2$ is sometimes a diameter (for $p=-12$), but not always a diameter.

Therefore Statement-1 is True.

4. Condition for $L_1$ to be a diameter

A line is a diameter of the circle if it passes through the centre. For $L_1$, $d_1=0 \Rightarrow p+6=0 \Rightarrow p=-6$

Now check $L_2$ for $p=-6$: $L_2: 2x+3y-3=0$ Its distance from centre is $d_2=\frac{|-6+15-3|}{\sqrt{13}}=\frac{6}{\sqrt{13}}$ Since $\frac{6}{\sqrt{13}}<2$ this line intersects the circle in two points, hence it is a chord.

So when $L_1$ is a diameter, $L_2$ is a chord.

Therefore Statement-2 is False.

5. Choose the correct option

• Statement-1: True
• Statement-2: False

Hence the correct option is: $\boxed{\text{C}}$

6. Comparison with stored answer

Stored correct answer: $\text{C}$

Our derived answer matches the stored answer.