Solutions to this question by accurate drawing will not be accepted. $OABC$ is a trapezium with right angle $OAB$. The coordinates of $A$ and $B$ are $(-2k, 3k)$ and $(0, t)$ respectively, where $k > 0$, and $OA = \sqrt{117}$ units. Find

1. Set up an equation using the given length of $OA$.

Using the distance formula for $O(0,0)$ and $A(-2k, 3k)$:
$OA^2 = (-2k - 0)^2 + (3k - 0)^2$
$(-2k)^2 + (3k)^2 = (\sqrt{117})^2$
$4k^2 + 9k^2 = 117$
$13k^2 = 117$

2. Solve for $k$ and find the coordinates of $A$.

$k^2 = \frac{117}{13} = 9$
Since $k > 0$, $k = 3$.
Substitute $k = 3$ into the coordinates of $A$:
$A = (-2(3), 3(3)) = (-6, 9)$ (shown)

1. Find the gradient of the line segment $OA$.

Gradient of $OA = \frac{9 - 0}{-6 - 0} = -\frac{3}{2}$

2. Find the gradient of $AB$ using the perpendicularity condition.

Since $\angle OAB = 90^\circ$, $OA \perp AB$.
Gradient of $AB = -\frac{1}{\text{Gradient of } OA} = \frac{2}{3}$

3. Find the equation of the line $AB$.

Using point $A(-6, 9)$ and gradient $\frac{2}{3}$:
$y - 9 = \frac{2}{3}(x - (-6))$
$y - 9 = \frac{2}{3}(x + 6)$
$y - 9 = \frac{2}{3}x + 4$
$y = \frac{2}{3}x + 13$

1. Find the coordinates of $B$ using the equation of $AB$ from part (b)

From part (a), $A$ is $(-6, 9)$.
Gradient of $OA = \frac{9 - 0}{-6 - 0} = -\frac{3}{2}$.
Since $OAB$ is a right angle, $AB \perp OA$, so gradient of $AB = \frac{2}{3}$.
Equation of $AB$: $y - 9 = \frac{2}{3}(x - (-6)) \implies y = \frac{2}{3}x + 13$.
Since $B$ lies on the $y$-axis, its $x$-coordinate is $0$, so $B$ is $(0, 13)$.

2. Determine the vector $\vec{AB}$

$\vec{AB} = \begin{pmatrix} 0 - (-6) \\ 13 - 9 \end{pmatrix} = \begin{pmatrix} 6 \\ 4 \end{pmatrix}$

3. Use the parallel property and length ratio to find the coordinates of $C$

Since $OABC$ is a trapezium with $OC \parallel AB$ and $C$ is in the first quadrant, $\vec{OC}$ is in the same direction as $\vec{AB}$.
Given $OC = \frac{1}{2}AB$, we have:
$\vec{OC} = \frac{1}{2}\vec{AB} = \frac{1}{2}\begin{pmatrix} 6 \\ 4 \end{pmatrix} = \begin{pmatrix} 3 \\ 2 \end{pmatrix}$
Since $O$ is the origin $(0,0)$, the coordinates of $C$ are $(3, 2)$.

1. State the coordinates of the vertices of the trapezium in order

From earlier parts, the coordinates are $O(0, 0)$, $C(3, 2)$, $B(0, 13)$, and $A(-6, 9)$.
Arranging them in an anti-clockwise direction: $O \rightarrow C \rightarrow B \rightarrow A \rightarrow O$.

2. Apply the shoelace formula to calculate the area

$\begin{aligned} \text{Area} &= \frac{1}{2} \left| \begin{matrix} 0 & 3 & 0 & -6 & 0 \\ 0 & 2 & 13 & 9 & 0 \end{matrix} \right| \\ &= \frac{1}{2} | (0)(2) + (3)(13) + (0)(9) + (-6)(0) - (0)(3) - (2)(0) - (13)(-6) - (9)(0) | \\ &= \frac{1}{2} | 0 + 39 + 0 + 0 - 0 - 0 + 78 - 0 | \\ &= \frac{1}{2} (117) \\ &= 58.5 \end{aligned}$

1. Explain why OB is the diameter using circle properties

It is given that $\angle OAB = 90^\circ$. Since triangle $OAB$ lies in the circle, the angle subtended by the chord $OB$ at the circumference is $90^\circ$. By the converse of the angle in a semicircle theorem, $OB$ must be the diameter of the circle.

1. Find the coordinates of $B$ using the fact that $AB$ is perpendicular to $OA$

Gradient of $OA = \frac{9 - 0}{-6 - 0} = -\frac{3}{2}$
Since $\angle OAB = 90^\circ$, gradient of $AB = \frac{2}{3}$.
Equation of $AB$: $y - 9 = \frac{2}{3}(x - (-6)) \implies y = \frac{2}{3}x + 13$
Since $B$ lies on the $y$-axis, $x = 0$, so $y = 13$. Thus, $B$ is $(0, 13)$.

2. Find the center and radius of the circle

Since $\angle OAB = 90^\circ$, $OB$ is the diameter of the circle.
Center of circle = midpoint of $OB = \left(\frac{0+0}{2}, \frac{0+13}{2}\right) = (0, 6.5)$
Radius = $\frac{13}{2} = 6.5$

3. Form the equation of the circle

Equation of circle: $(x - 0)^2 + (y - 6.5)^2 = 6.5^2$
$x^2 + y^2 - 13y + 42.25 = 42.25$
$x^2 + y^2 - 13y = 0$