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(D) Let the equation of the circle be $x^2+y^2+2gx+2fy+c=0$.Since it passes through $(1, -2)$,we have $1+4+2g-4f+c=0 \Rightarrow 2g-4f+c=-5$ $(i)$.Sin

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$x^2+y^2-6x+2y+5=0$

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(D) Let the equation of the circle be $x^2+y^2+2gx+2fy+c=0$.Since it passes through $(1, -2)$,we have $1+4+2g-4f+c=0 \Rightarrow 2g-4f+c=-5$ $(i)$.Since it passes through $(4, -3)$,we have $16+9+8g-6f+c=0 \Rightarrow 8g-6f+c=-25$ $(ii)$.Subtracting $(i)$ from $(ii)$: $(8g-6f+c) - (2g-4f+c) = -25 - (-5) \Rightarrow 6g-2f = -20$ $(iii)$.The centre $(-g, -f)$ lies on $3x+2y=7$,so $3(-g)+2(-f)=7 \Rightarrow -3g-2f=7$ $(iv)$.Subtracting $(iv)$ from $(iii)$: $(6g-2f) - (-3g-2f) = -20 - 7$ $\Rightarrow 9g = -27$ $\Rightarrow g = -3$.Substituting $g=-3$ in $(iii)$: $6(-3)-2f = -20$ $\Rightarrow -18-2f = -20$ $\Rightarrow -2f = -2$ $\Rightarrow f = 1$.Substituting $g=-3$ and $f=1$ in $(i)$: $2(-3)-4(1)+c = -5$ $\Rightarrow -6-4+c = -5$ $\Rightarrow c = 5$.The equation of the circle is $x^2+y^2-6x+2y+5=0$.

The equation of the circle passing through the points $(1, -2)$ and $(4, -3)$ and whose centre lies on the line $3x + 2y = 7$ is

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