Finding area of a triangle using equation of a circle

Question

**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 $OBC$

Answer 1

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.

Answer 2

Some hints:

1. Using the $y$ coordinate of $A$, find the $x$ coordinate of $A$.
2. Find the slope of line $OA$.
3. From that, calculate the slope of the tangent line $BC$.
4. Using point-slope form, calculate the intercepts.
5. Profit!

Answer 3

Just going off the question alone and assuming the diagram is not drawn to scale, have some good information to off of.

Knowing the $y$ coordinate is $-1$, we can plug that into the equation of the circle to get the $x$ coordinate:

$x^2 + (-1)^2 = 5 \implies x=2$

This gives us a slope of the line $OA$ to be $\frac{-1}{2}$ and the tangent slope to be $2$. Therefore, we have a line that passes through $(2,-1)$ with a slope of $2$. We can get the equation using:

$ y=mx+b \implies -1 = 2 \cdot 2+b \implies b = -5$

Our $y$ intercept is -5 (height), and our $x$ intercept is then $0=2x-5 \implies x=2.5$ (width). Our area is:

$ \frac{5\cdot2.5}{2} = 6.25$