This problem requires finding the equation of a circle given information about its diameters and area. The key steps involve finding the circle's center by intersecting the diameter lines and determining the radius from the given area.

Finding the Circle Center

The center of the circle is the point where the two given diameters intersect. We need to solve the system of linear equations:

• Diameter 1: $2x - 3y + 12 = 0$
• Diameter 2: $x + 4y - 5 = 0$

To solve this system, we can use substitution or elimination. Let's use elimination:

1. Multiply the second equation by 2 to match the coefficient of $x$ in the first equation: $2 \times (x + 4y - 5) = 0 \implies 2x + 8y - 10 = 0$
2. Now, subtract the first equation ($2x - 3y + 12 = 0$) from this new equation ($2x + 8y - 10 = 0$):

   $$(2x + 8y - 10) - (2x - 3y + 12) = 0$$ $$2x + 8y - 10 - 2x + 3y - 12 = 0$$ $$(8y + 3y) + (-10 - 12) = 0$$ $$11y - 22 = 0$$ $$11y = 22$$ $$y = \frac{22}{11}$$ $$y = 2$$

3. Substitute the value of $y = 2$ back into the second original equation ($x + 4y - 5 = 0$): $$x + 4(2) - 5 = 0$$ $$x + 8 - 5 = 0$$ $$x + 3 = 0$$ $$x = -3$$

Therefore, the center of the circle, $(h, k)$, is $(-3, 2)$.

Calculating the Radius Squared

The area of the circle is given as 154 square units. The formula for the area ($A$) of a circle with radius ($r$) is $A = \pi r^2$. We can use the approximation $\pi \approx \frac{22}{7}$.

$$A = \pi r^2$$

$$154 = \frac{22}{7} r^2$$

To find $r^2$, rearrange the equation:

$$r^2 = \frac{154 \times 7}{22}$$

Simplify the fraction:

$$r^2 = \frac{(7 \times 22) \times 7}{22}$$

$$r^2 = 7 \times 7$$

$$r^2 = 49$$

So, the radius squared ($r^2$) is 49.

Deriving the Circle Equation

The standard equation of a circle with center $(h, k)$ and radius $r$ is:

$$(x - h)^2 + (y - k)^2 = r^2$$

Substitute the center $(h, k) = (-3, 2)$ and $r^2 = 49$ into the standard equation:

$$(x - (-3))^2 + (y - 2)^2 = 49$$

$$(x + 3)^2 + (y - 2)^2 = 49$$

Now, expand the squared terms:

$$(x^2 + 2 \cdot x \cdot 3 + 3^2) + (y^2 - 2 \cdot y \cdot 2 + 2^2) = 49$$

$$(x^2 + 6x + 9) + (y^2 - 4y + 4) = 49$$

Combine the terms:

$$x^2 + y^2 + 6x - 4y + 9 + 4 = 49$$

$$x^2 + y^2 + 6x - 4y + 13 = 49$$

Move the constant term to the left side:

$$x^2 + y^2 + 6x - 4y + 13 - 49 = 0$$

$$x^2 + y^2 + 6x - 4y - 36 = 0$$

Comparing the Result with Options

The derived equation of the circle is $x^2 + y^2 + 6x - 4y - 36 = 0$. Let's compare this with the given options:

• Option 1: $x^2 + y^2 + 6x - 4y - 36 = 0$
• Option 2: $x^2 + y^2 + 6x + 4y - 36 = 0$
• Option 3: $x^2 + y^2 - 6x + 4y + 25 = 0$

The derived equation matches Option 1.