The equation of the circle having centre (1, | vedclass

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(A) First,find the point of intersection of the lines $3x + y = 14$ and $2x + 5y = 18$.Multiplying the first equation by $5$,we get $15x + 5y = 70$.Su

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${x^2} + {y^2} - 2x + 4y - 20 = 0$

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(A) First,find the point of intersection of the lines $3x + y = 14$ and $2x + 5y = 18$.Multiplying the first equation by $5$,we get $15x + 5y = 70$.Subtracting the second equation from this,we get $(15x - 2x) = 70 - 18$,which gives $13x = 52$,so $x = 4$.Substituting $x = 4$ into $3x + y = 14$,we get $3(4) + y = 14$,so $y = 2$.The point of intersection is $(4, 2)$.The radius $r$ is the distance between the center $(1, -2)$ and the point $(4, 2)$:$r^2 = (4 - 1)^2 + (2 - (-2))^2 = 3^2 + 4^2 = 9 + 16 = 25$.The equation of the circle with center $(h, k)$ and radius $r$ is $(x - h)^2 + (y - k)^2 = r^2$.Substituting $(h, k) = (1, -2)$ and $r^2 = 25$,we get $(x - 1)^2 + (y + 2)^2 = 25$.Expanding this,$x^2 - 2x + 1 + y^2 + 4y + 4 = 25$,which simplifies to $x^2 + y^2 - 2x + 4y - 20 = 0$.

The equation of the circle having centre $(1, -2)$ and passing through the point of intersection of lines $3x + y = 14$ and $2x + 5y = 18$ is

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