Ans: (a) $\Delta DBC \sim \Delta DOA$ (b) (i) $(0,4)$ (ii) $(x-3)^2+(y-2)^2=13$
- $\Delta DBC \sim \Delta DOA$.
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- Let $C=(0,y)$. Since $\Delta DBC \sim \Delta DOA$, then
$\begin{array}{rcl}
\dfrac{\mbox{area of }\Delta OAD}{\mbox{area of }\Delta BCD} & = & \left(\dfrac{AD}{CD}\right)^2 \\
\dfrac{45}{16} & = & \left(\dfrac{\sqrt{180}}{12-y}\right)^2 \\
y^2 -24y+80 & = & 0 \\
(y-20)(y-4) & = & 0
\end{array}$Therefore $y=20$ or $y=4$. Since $C$ is below $D$, then the coordinates of $C$ is $(0,4)$.
- Since $\angle AOD=90^\circ$, $AC$ is a diameter. Therefore, the mid-point of $AC$ is the centre of the circle.
The centre
$\begin{array}{cl}
= & \left( \dfrac{0+6}{2},\dfrac{4+0}{2} \right) \\
= & (3,2)
\end{array}$The radius
$\begin{array}{cl}
= & \sqrt{(3-0)^2+(2-0)^2} \\
= & \sqrt{13}
\end{array}$Therefore, the equation of the circle is $(x-3)^2+(y-2)^2 = 13$.
- Let $C=(0,y)$. Since $\Delta DBC \sim \Delta DOA$, then
