The radius of the circle circumscribing the quadrilateral formed by the lines $7 x + y - 57 = 0, 4 x - 3 y - 29 = 0, 7 x + y - 7 = 0$ and $3 x + 4 y - 3 = 0$ in order is

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Solution

Let the equation of the circumcircie of the quadrilateral be
$(7 x + y - 57) (7 x + y - 7) + λ (4 x - 3 y - 29) (3 x + 4 y - 3) = 0$
$∴$ coefficient of $x^{2} =$ coefficient of $y^{2}$
$\Rightarrow 49 + 12 λ = 1 - 12 λ \Rightarrow λ = - 2$
$∴$ equation (1) becomes 25 $(x^{2} + y^{2}) - 250 x + 150 y + 225 = 0$
i.e., $x^{2} + y^{2} - 10 x + 6 y + 9 = 0$
Hence the radius = $\sqrt{5^{2} + (- 3)^{2} - 9} = 5$

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The radius of the circle circumscribing the quadrilateral formed by the lines 7x+y−57=0,4x−3y−29=0,7x+y−7=0 and 3x+4y−3=0 in order is

The radius of the circle circumscribing the quadrilateral formed by the lines $7 x + y - 57 = 0, 4 x - 3 y - 29 = 0, 7 x + y - 7 = 0$ and $3 x + 4 y - 3 = 0$ in order is