Question

Any circle through the point of intersection of the lines $x+\sqrt{3}y=1$ and $\sqrt{3}x-y=2$ if intersects these lines at points $P$ and $Q$, then the angle subtended by the are $PQ$ at its centre is

• A. ${180}^o$
• B. ${90}^o$
• C. ${120}^o$
• D. Depends on centre and radius

Answer 1

Answer: (A)

${180}^o$

Let the point of intersection of two lines is $A$.

$\therefore$ The angle subtended by $PQ$ on centre $C$ $=2\times$ the angle subtended by $PQ$ on point $A$.

For $x+\sqrt{3}y=1,m_1=\frac{-1}{\sqrt{3}}$ and for $\sqrt{3}x-y=2,m_2=\sqrt{3}$

$\because m_1\times m_2=\frac{-1}{\sqrt{3}}\times \sqrt{3}=-1$,

$\therefore \angle A={90}^o$

$\therefore$ The angle subtended by are PQ at its centre $=2\times {90}^o={180}^o$