Finding area of a triangle using equation of a circle **Ignore notes I made they are stupid
Without a calculator
Question reads:

The diagram shows a sketch of the circle with equation $x^2 + y^2 = 5$.
  The $y$-coordinate of point $A$ is $-1$.
  The tangent to the circle at $A$ crosses the axes at $B$ and $C$ as shown.
Find the area of triangle $OBC$



The circle has radius $\sqrt{5}$ and this is the length of $OA$. Call the projection of the point A on the x-axis A’. You now have that the angle between A’ and A equals $\arcsin(\frac{OA’}{OA}) = \theta$. You can now calculate the area with $$ 0.5 \frac{\sqrt{5}}{\cos(\theta)} \frac{\sqrt{5}}{\cos(0.5\pi -\theta)}$$
Because of the tangent line the triangles OAB and OAC are right angled.
