Animated Solution for Mathematics - Circles: 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

We are given two lines: $3x-4y-7=0$ and $2x-3y-5=0$. These lines represent diameters of the circle.

By definition, a diameter must pass through the center of the circle. Therefore, the intersection point of these two diameters is the center of the circle, denoted as $C(h,k)$.

To find the intersection, we solve the system of linear equations:

Using the elimination method, we multiply the first equation by $3$ and the second by $4$:

Subtracting the second equation from the first, the $y$ terms cancel out, yielding:

Substituting $x=1$ into the equation $2x-3y=5$:

Thus, the center of the circle is $C(1,-1)$.

The problem states that the area of the circle is $49\pi$. We use the standard area formula:

Dividing both sides by $\pi$, we obtain:

The standard form of a circle's equation is $(x-h{)}^2+(y-k{)}^2=r^2$. Substituting our center $(1,-1)$ and $r^2=49$:

Expanding the squares, we get:

Combining the constant terms:

Subtracting $49$ from both sides, we arrive at the final equation:

$$x^2+y^2-2x+2y-47=0$$