If the lines $$3x - 4y - 7 = 0$$ and $$2x - 3y - 5 = 0$$ are two diameters of a circle of area $$49\pi$$ square units, the equation of the circle is

The centre of a circle lies on every diameter. Hence the point of intersection of the two given diameter-lines is the centre of the required circle.

Find this intersection.
Solve the simultaneous equations $$3x - 4y - 7 = 0 \; -(1)$$ $$2x - 3y - 5 = 0 \; -(2)$$

From $$(2)$$, write $$2x = 3y + 5$$, so $$x = \frac{3y + 5}{2}$$ and substitute in $$(1)$$:

$$3\left(\frac{3y + 5}{2}\right) - 4y - 7 = 0$$ $$\Rightarrow \frac{9y + 15}{2} - 4y - 7 = 0$$ $$\Rightarrow 9y + 15 - 8y - 14 = 0$$ $$\Rightarrow y = -1$$

Put $$y = -1$$ in $$(2)$$: $$2x - 3(-1) - 5 = 0 \;\Rightarrow\; 2x + 3 - 5 = 0 \;\Rightarrow\; 2x = 2 \;\Rightarrow\; x = 1$$

Thus the centre is $$C(1,-1)$$.

The area of the circle is given as $$49\pi$$, so $$\pi r^2 = 49\pi \;\Rightarrow\; r^2 = 49 \;\Rightarrow\; r = 7$$.

The equation of a circle with centre $$(h,k)$$ and radius $$r$$ is $$(x - h)^2 + (y - k)^2 = r^2$$.

Substituting $$h = 1,\, k = -1,\, r^2 = 49$$: $$(x - 1)^2 + (y + 1)^2 = 49$$

Expand: $$x^2 - 2x + 1 + y^2 + 2y + 1 = 49$$ $$\Rightarrow x^2 + y^2 - 2x + 2y + 2 - 49 = 0$$ $$\Rightarrow x^2 + y^2 - 2x + 2y - 47 = 0$$

Hence the equation of the circle is $$x^2 + y^2 - 2x + 2y - 47 = 0$$.

Option D which is: $$x^2 + y^2 - 2x + 2y - 47 = 0$$