JEE Advanced · Mathematics · 9. Straight Lines
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 \(C\).
- A
Statement 1 is true, Statement 2 is true, Statement 2 is a correct explanation for Statement 1.
- B
Statement 1 is true, Statement 2 is true, Statement 2 is not a correct explanation for Statement 1.
- C
Statement 1 is true, Statement 2 is false.
- D
Statement 1 is false, Statement 2 is true
Answer & Solution
Correct Answer
(C)
Statement 1 is true, Statement 2 is false.
Step-by-step Solution
Detailed explanation
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.
\[
\begin{aligned}
(x+3)^2+(y-5)^2 & =9+25-30=4 \\
\Rightarrow \quad \quad(x+3)^2+(y-5)^2 & =2^2 \\
\text { Centre } & =(3,-5)
\end{aligned}
\]
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.
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