If the lines x+2y-5=0 and 2x-3y+4=0 lie along | vedclass

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(A) Given: The lines $x+2y-5=0$ and $2x-3y+4=0$ are diameters of the circle. Since the intersection of diameters is the center of the circle,we solve

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

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(A) Given: The lines $x+2y-5=0$ and $2x-3y+4=0$ are diameters of the circle. Since the intersection of diameters is the center of the circle,we solve the system of equations: $x+2y=5$ $(i)$ $2x-3y=-4$ (ii) Multiplying $(i)$ by $2$,we get $2x+4y=10$ (iii). Subtracting (ii) from (iii): $(2x+4y)-(2x-3y) = 10-(-4)$ $\Rightarrow 7y=14$ $\Rightarrow y=2$. Substituting $y=2$ into $(i)$: $x+2(2)=5 \Rightarrow x=1$. Thus,the center $(h, k)$ is $(1, 2)$. The area of the circle is $9\pi$,so $\pi r^2 = 9\pi \Rightarrow r^2=9$. The equation of the circle is $(x-h)^2 + (y-k)^2 = r^2$. $(x-1)^2 + (y-2)^2 = 9 (x^2-2x+1) + (y^2-4y+4) = 9 x^2+y^2-2x-4y-4=0$.

If the lines $x+2y-5=0$ and $2x-3y+4=0$ lie along diameters of a circle of area $9\pi$,then the equation of the circle is

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