Question

Prove that the equation to the circle described on the straight line joining the points $(1,{60}^{\circ })$ and $(2,{30}^{\circ })$ as diameter is $r^2-r\left[\cos \right(\theta -{20}^{\circ })+2\cos (\theta -{30}^{\circ }\left)\right]+\sqrt{3}=0$.

Answer 1

If $(a,\alpha )$ and $(b,\beta )$ are the end points of diameter, the equation of circle is

$r^2-r\{a\cos (\theta -\alpha )+b\cos (\theta -\beta )\}+ab\cos (\alpha -\beta )=0$

Substituting values of given coordinates, we get

$r^2-r\{1.\cos (\theta -{60}^{\circ })+2.\cos (\theta -{30}^{\circ })\}+\left(1\right)\left(2\right)\cos ({60}^{\circ }-{30}^{\circ })=0r^2-r\{\cos (\theta -{60}^{\circ })+2\cos (\theta -{30}^{\circ })\}+2\times \frac{\sqrt{3}}{2}=0r^2-r\{\cos (\theta -{60}^{\circ })+2\cos (\theta -{30}^{\circ })\}+\sqrt{3}=0$

Hence proved.