Question

The equation of circle described on the straight line joining the points $(4,\frac{\pi }{6})$ and $(3,\frac{\pi }{3})$ as diameter is

• A. $r^2+r[4\cos (\theta -\frac{\pi }{6})+3\cos (\theta -\frac{\pi }{3})]+6\sqrt{3}=0$
• B. $r^2-r[4\cos (\theta -\frac{\pi }{6})+3\cos (\theta -\frac{\pi }{3})]+6\sqrt{3}=0$
• C. $r^2+r[4\cos (\theta -\frac{\pi }{6})-3\cos (\theta -\frac{\pi }{3})]+6\sqrt{3}=0$
• D. $r^2-r[4\cos (\theta -\frac{\pi }{6})+3\cos (\theta -\frac{\pi }{3})]-6\sqrt{3}=0$

Answer 1

The correct option is B $r^2-r[4\cos (\theta -\frac{\pi }{6})+3\cos (\theta -\frac{\pi }{3})]+6\sqrt{3}=0$

$\therefore (AP{)}^2=4^2+r^2-8r\cos (\theta -\frac{\pi }{6})(BP{)}^2=3^2+r^2-6r\cos (\frac{\pi }{3}-\theta )and(AB{)}^2=3^2+4^2-24\cos (\frac{\pi }{3}-\frac{\pi }{6})=25-24\cos \left(\cos \frac{\pi }{6}\right)=25-24\times \frac{\sqrt{3}}{2}=25-12\sqrt{3}$
$\because \angle APB=90(\because$Angle in a semicircle is a right angle)
$\therefore (AP{)}^2+(BP{)}^2=(AB{)}^2\Rightarrow 2r^2+25-2r[4\cos (\theta -\frac{\pi }{6})+3\cos (\frac{\pi }{3}-\theta )]=25-12\sqrt{3}{r}^2-r[4\cos (\theta -\frac{\pi }{6})+3\cos (\theta -\frac{\pi }{3})]+6\sqrt{3}=0$