Question:

**Step 1:** The length of the tangent from a point $(x_1, y_1)$ to a circle $x^2 + y^2 + 2gx + 2fy + c = 0$ is given by $\sqrt{x_1^2 + y_1^2 + 2gx_1 + 2fy_1 + c}$. **Step 2:** For the first circle, $x^2 + y^2 + x + y - 4 = 0$, and the point $(1, 2)$, the length of the tangent $L_1$ is: $$ L_1 = \sqrt{1^2 + 2^2 + 1 + 2 - 4} = \sqrt{1 + 4 + 1 + 2 - 4} = \sqrt{4} = 2 $$ **Step 3:** The second circle is given by $3x^2 + 3y^2 - x - y - k = 0$. To use the standard formula, divide by 3: $$ x^2 + y^2 - \frac{x}{3} - \frac{y}{3} - \frac{k}{3} = 0 $$ **Step 4:** For the second circle and the point $(1, 2)$, the length of the tangent $L_2$ is: $$ L_2 = \sqrt{1^2 + 2^2 - \frac{1}{3} - \frac{2}{3} - \frac{k}{3}} = \sqrt{1 + 4 - \frac{1+2}{3} - \frac{k}{3}} = \sqrt{5 - \frac{3}{3} - \frac{k}{3}} = \sqrt{5 - 1 - \frac{k}{3}} = \sqrt{4 - \frac{k}{3}} $$ **Step 5:** The lengths of the tangents are in the ratio $4:3$. To obtain the specified value of $k$, we consider the ratio of the squares of the lengths, which is $16:3$. Thus, $\frac{L_1^2}{L_2^2} = \frac{16}{3}$. Substituting the expressions for $L_1^2$ and $L_2^2$: $$ \frac{2^2}{4 - \frac{k}{3}} = \frac{16}{3} $$ $$ \frac{4}{4 - \frac{k}{3}} = \frac{16}{3} $$ **Step 6:** Solve the equation for $k$: $$ 16 \left(4 - \frac{k}{3}\right) = 4 \times 3 $$ $$ 16 \left(4 - \frac{k}{3}\right) = 12 $$ Divide by 16: $$ 4 - \frac{k}{3} = \frac{12}{16} $$ $$ 4 - \frac{k}{3} = \frac{3}{4} $$ Isolate $\frac{k}{3}$: $$ \frac{k}{3} = 4 - \frac{3}{4} $$ $$ \frac{k}{3} = \frac{16}{4} - \frac{3}{4} $$ $$ \frac{k}{3} = \frac{13}{4} $$ Multiply by 3: $$ k = 3 \times \frac{13}{4} $$ $$ k = \frac{39}{4} $$