The equation of the circle having center $(1, - 2)$ and passes through the point of intersection of lines $3 x + y = 14, 2 x + y = 18.$

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Solution

$L_{1} : 3 x + y = 14, L_{2} : 2 x + y = 18$
Intersection pt of $L_{1}$ and $L_{2}$
$3 x + y = 14$
$2 x + y = 18$
$-$ $-$ $-$
$¯ ¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯ ¯ x = - 4$
$3 (- 4) + y = 14$
$y = 14 + 12$
$y = 26$
$A (- 4, 26), O (1, - 2)$
$r = O A = \sqrt{(5)^{2} + (28)^{2}} = \sqrt{809}$
$r = 28.44 c m$
equation of circle : $(x - h)^{2} + (y - k)^{2} = r^{2}$
$(x - 1)^{2} + (y - 2)^{2} = 809$
$x^{2} + 1 - 2 x + y^{2} + 4 + 2 y = 809$
$x^{2} + y^{2} - 2 x + 2 y - 804 = 0$

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