A point moving around circle $(x + 4)^{2} + (y + 2)^{2} = 25$ with centre C broke away from it either at the point A or point B on the circle and moved along a tangent to the circle passing through the point D(3, -3). Find the following.
(i) Equation of the tangents at A and B
(ii) Coordinates of the points A and B
(iii) Angle ADB and the maximum and minimum distances of the point D from the circle.
(iv) Area of quadrilateral ABCD and the $△ D A B$
(v) Equation of the circle circumscribing the $△ D A B$ and also the intercepts made by this circle on the coordinate axes.

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Solution

$(i)$
$\begin{matrix} C D = \sqrt{45 + 1} = \sqrt{50} A D = \sqrt{C D^{2} + A C^{2}} = \sqrt{50 - 25} = 5 H e n c e A C = B C = B D = A D = 5 ∠ A D C = ∠ B D C = 45^{\circ} L e t S l o p e o f A D = m_{1} S l o p e o f B D = m_{2} S l o p e o f C D = \frac{- 2 + 3}{- 4 - 3} = \frac{- 1}{7} \frac{π}{4} = ∣ ∣ ∣ \frac{m + \frac{1}{7}}{1 - m \frac{1}{7}} ∣ ∣ ∣ \frac{π}{4} = ∣ ∣ ∣ \frac{m + \frac{1}{7}}{1 - \frac{m}{7}} ∣ ∣ ∣ T a k i n g + v e & T a k i n g - v e (1 - \frac{m}{7}) = (\frac{1}{7} + m) & - 1 + \frac{m}{7} = m + \frac{1}{7} m = \frac{3}{4} & m = \frac{- 4}{3} E q u a t i o n o f l i n e & E q u a t i o n o f l i n e y + 3 = \frac{3}{4} (x - 3) & y + 3 = \frac{- 4}{3} (x - 3) 4 y - 3 x + 21 = 0 & 3 y + 4 x - 3 = 0 \end{matrix}$

$(i i)$
$\begin{matrix} S l o p e o f D C = \frac{- 1}{s l o p e o f A D} & S l o p e o f B C = \frac{- 1}{S t a g e o f B D} = \frac{- 4}{3} & = \frac{3}{4} E q u a t i o n o f A C & E q u a t i o n o f B C y + 2^{\circ} = \frac{- 4}{3} (x + 4) & y + 2 = \frac{3}{4} (x + 4) 3 y + 6 = - 4 x - 16 & 4 y + 8 = 3 x + 12 4 x + 3 y + 22 = 0 & 3 x - 4 y + 4 = 0 \end{matrix}$
For point $A$ , intersection of $A C$ and $A D$
For point $B$ , intersection of $B C$ and $B D$
Point $A = (0, 1)$
$B = (- 1, - 6)$

$(i i i)$
Minimum distance = $= 5 \sqrt{2} - 5$
Maximum distance = $= 5 \sqrt{2} + 5$

$(i v)$
Area of quadrilateral $A B B C = 5^{2} = 25$
Area of $O D A B = \frac{1}{2} \times 25 = 12.5$

$(v)$
Mid point of $C D = (\frac{- 1}{2}, \frac{- 5}{2})$
Equation of circle
$\begin{matrix} {(x + \frac{1}{2})}^{2} + {(y + \frac{5}{2})}^{2} = {(\frac{5 \sqrt{2}}{2})}^{2} x^{2} + y^{2} + x + 5 y - 6 = 0 \end{matrix}$

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