Circle Ques 117
- Consider
$$ \begin{aligned} & L _1: 2 x+3 y+p-3=0 \\ & L _2: 2 x+3 y+p+3=0 \end{aligned} $$
where, $p$ is a real number and
$$ C: x^{2}+y^{2}-6 x+10 y+30=0 $$
Statement I If line $L _1$ is a chord of circle $C$, then line $L _2$ is not always a diameter of circle $C$.
Statement II If line $L _1$ is a diameter of circle $C$, then line $L _2$ is not a chord of circle $C$.
$(2008,3 M)$
Fill in the Blanks
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Answer:
Correct Answer: 117.(c)
Solution:
Formula:
- Equation of given circle $C$ is
$$ \begin{array}{ll} & (x-3)^{2}+(y+5)^{2}=9+25-30 \\ \text { i.e. } \quad & (x-3)^{2}+(y+5)^{2}=2^{2} \end{array} $$
Centre $=(3,-5)$
If $L _1$ is diameter, then $2(3)+3(-5)+p-3=0 \Rightarrow p=12$
$$ \therefore \quad \begin{aligned} L _1 \text { is } 2 x+3 y+9 & =0 \\ L _2 \text { is } 2 x+3 y+15 & =0 \end{aligned} $$
Distance of centre of circle from $L _2$ equals
$$ \left|\frac{2(3)+3(-5)+15}{\sqrt{2^{2}+3^{2}}}\right|=\frac{6}{\sqrt{13}}<2 $$
[radius of circle]
$\therefore \quad L _2$ is a chord of circle $C$.
Statement II is false.
BETA
