Question:

Step 1: Determine the circle's properties and the length of the tangent $PT$. The equation of the circle is given as $x^2 + y^2 + 4x - 6y - 3 = 0$. The center of the circle $C$ is found by comparing with the general form $x^2 + y^2 + 2gx + 2fy + c = 0$, so $C(-g, -f) = (-4/2, -(-6)/2) = (-2, 3)$. The radius of the circle $r$ is $\sqrt{g^2 + f^2 - c} = \sqrt{(-2)^2 + (3)^2 - (-3)} = \sqrt{4 + 9 + 3} = \sqrt{16} = 4$. Thus, $CT = CB = r = 4$. The point $P$ is given as $(5, 3 + \sqrt{3})$. The length of the tangent $PT$ from an external point $P(x_1, y_1)$ to the circle $S=0$ is given by $PT = \sqrt{S_1}$. $$ S_1 = x_1^2 + y_1^2 + 4x_1 - 6y_1 - 3 $$ $$ PT = \sqrt{5^2 + (3+\sqrt{3})^2 + 4(5) - 6(3+\sqrt{3}) - 3} $$ $$ PT = \sqrt{25 + (9 + 6\sqrt{3} + 3) + 20 - (18 + 6\sqrt{3}) - 3} $$ $$ PT = \sqrt{25 + 12 + 6\sqrt{3} + 20 - 18 - 6\sqrt{3} - 3} $$ $$ PT = \sqrt{(25 + 12 + 20 - 18 - 3) + (6\sqrt{3} - 6\sqrt{3})} = \sqrt{36} = 6 $$ Step 2: Establish the geometry related to the normal $BT$ and $\triangle PBT$. The problem states that $BT$ is the normal at $T$. A normal to a circle passes through its center. Therefore, points $B, C, T$ are collinear. This implies that $BT$ is a diameter of the circle. Hence, the length of $BT = 2r = 2(4) = 8$. Since $PT$ is a tangent to the circle at $T$ and $BT$ is a normal (diameter) through $T$, the angle $\angle PTB$ must be $90^\circ$. Thus, $\triangle PBT$ is a right-angled triangle with the right angle at $T$. In $\triangle PBT$, we have $PT = 6$ and $BT = 8$. Step 3: Determine the value of $\sin \alpha$. Let $\angle ACB = \alpha$. The original solution uses the relation $\tan(\alpha/2) = 8/6$. This implies that $\alpha/2$ is the angle $\angle BPT$ in the right-angled triangle $\triangle PBT$. In $\triangle PBT$, the tangent of $\angle BPT$ is the ratio of the side opposite to $\angle BPT$ ($BT$) to the side adjacent to $\angle BPT$ ($PT$). $$ \tan\left(\frac{\alpha}{2}\right) = \tan(\angle BPT) = \frac{BT}{PT} = \frac{8}{6} = \frac{4}{3} $$ Now, we calculate $\sin \alpha$ using the double angle formula $\sin \alpha = \frac{2 \tan(\alpha/2)}{1 + \tan^2(\alpha/2)}$: $$ \sin \alpha = \frac{2\left(\frac{4}{3}\right)}{1 + \left(\frac{4}{3}\right)^2} = \frac{\frac{8}{3}}{1 + \frac{16}{9}} = \frac{\frac{8}{3}}{\frac{9+16}{9}} = \frac{\frac{8}{3}}{\frac{25}{9}} $$ $$ \sin \alpha = \frac{8}{3} \times \frac{9}{25} = \frac{24}{25} $$ Step 4: Calculate the areas of $\triangle CAB$ and $\triangle CAT$. For $\triangle CAB$: The sides $CA$ and $CB$ are both radii of the circle, so $CA = CB = r = 4$. The angle between them is $\angle ACB = \alpha$. $$ Ar(\triangle CAB) = \frac{1}{2} CA \cdot CB \sin(\angle ACB) = \frac{1}{2} (4)(4) \sin \alpha = 8 \sin \alpha $$ For $\triangle CAT$: The sides $CA$ and $CT$ are both radii of the circle, so $CA = CT = r = 4$. Since $B, C, T$ are collinear, the angle $\angle ACT$ is supplementary to $\angle ACB$. $$ \angle ACT = \pi - \angle ACB = \pi - \alpha $$ $$ Ar(\triangle CAT) = \frac{1}{2} CA \cdot CT \sin(\angle ACT) = \frac{1}{2} (4)(4) \sin(\pi - \alpha) $$ Since $\sin(\pi - \alpha) = \sin \alpha$: $$ Ar(\triangle CAT) = 8 \sin \alpha $$ Step 5: Calculate the total area and determine $k$. The sum of the areas is $Ar(\triangle CAB) + Ar(\triangle CAT)$: $$ Ar(\triangle CAB) + Ar(\triangle CAT) = 8 \sin \alpha + 8 \sin \alpha = 16 \sin \alpha $$ Substitute the value of $\sin \alpha = \frac{24}{25}$ from Step 3: $$ \text{Total Area} = 16 \left(\frac{24}{25}\right) = \frac{384}{25} $$ The problem states that $Ar(\triangle CAB) + Ar(\triangle CAT) = \frac{k}{25}$. By comparing, we find $k = 384$. Step 6: Calculate the final value using the Greatest Integer Function. We need to find the value of $(\sqrt{k} - 15)$, where $[.]$ denotes the Greatest Integer Function (G.I.F.). $$ (\sqrt{k} - 15) = (\sqrt{384} - 15) $$ To calculate the value, we estimate $\sqrt{384}$: Since $19^2 = 361$ and $20^2 = 400$, $\sqrt{384}$ is between 19 and 20. More precisely, $\sqrt{384} = \sqrt{64 \times 6} = 8\sqrt{6}$. Using $\sqrt{6} \approx 2.449$: $$ 8\sqrt{6} \approx 8 \times 2.449 = 19.592 $$ Now substitute this value into the expression: $$ \sqrt{384} - 15 \approx 19.592 - 15 = 4.592 $$ Applying the Greatest Integer Function: $$ [\sqrt{k} - 15] = [4.592] = 4 $$ The final answer is 4. <div class="key-concept"><strong>Key Concept:</strong> Use the tangent half-angle formula to relate the triangle dimensions and compute the combined area using sine addition.</div> <div class="trap-box"><strong>Trap:</strong> Forgetting that $\sin(\pi - \alpha) = \sin\alpha$ leads to miscalculation of the second triangle's area.</div>