The tangent at $\left(\alpha, \beta\right)$ to the circle $x^2 + y^2 = r^2$ cuts the axes of coordinates at $A$ and $B$. Prove that the area of $\triangle OAB = \dfrac{r^4}{2 \alpha \beta}$
Given point: $\;$ $\left(x_1, y_1\right) = \left(\alpha, \beta\right)$
Given circle: $\;$ $x^2 + y^2 = r^2$
Equation of tangent to the circle $\;$ $x^2 + y^2 = r^2$ $\;$ at the point $\;$ $\left(x_1, y_1\right)$ is $\;$ $x x_1 + yy_1 = r^2$
$\therefore \;$ Equation of tangent to the circle $\;$ $x^2 + y^2 = r^2$ $\;$ at the point $\;$ $\left(\alpha, \beta\right)$ $\;$ is
$\alpha x + \beta y = r^2$ $\;\;\; \cdots \; (1)$
Let equation $(1)$ cut the $X$ and the $Y$ axes at the points $A$ and $B$ respectively.
Then, $\;$ $A = \left(\dfrac{r^2}{\alpha}, 0\right)$, $\;$ $B = \left(0, \dfrac{r^2}{\beta}\right)$
$OA = \dfrac{r^2}{\alpha}$, $\;$ $OB = \dfrac{r^2}{\beta}$ $\;$ where $\;$ $O \left(0, 0\right)$ $\;$ is the origin
$\therefore \;$ Area of $\triangle OAB = \dfrac{1}{2} \times OA \times OB$
i.e. $\;$ Area of $\triangle OAB = \dfrac{1}{2} \times \dfrac{r^2}{\alpha} \times \dfrac{r^2}{\beta}$
i.e. $\;$ Area of $\triangle OAB = \dfrac{r^4}{2 \alpha \beta}$