Find the equation of the circle circumscribing the rectangle whose sides are
$x - 3 y = 4, 3 x + y = 22, x - 3 y = 14$ and $3 x + y = 62$ .

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Solution

The given equation are
$x - 3 y = 4 \dots (i) 3 x + y = 22 \dots (i i) x - 3 y = 14 \dots (i i i) 3 x + y = 62 \dots (i v)$
Let A, B, C and D are the points of intersection of die lines (i) and (ii), (ii) and (iii), (iii) and (iv), (iv) and (i)
$∴ A = (7, 1), B = (8, - 2), C = (20, 2)$ and $D = (19, 5)$
AC will be the diameter of the circle
So,
the equation of circle is
$(x - 7) (x - 20) + (y - 1) (y - 2) = 0$

$\Rightarrow x^{2} + y^{2} - 27 x - 3 y + 142 = 0$

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Determine graphically the vertices of the triangle, the equations of whose sides are given below:

(i) 2y - x = 8, 5y-x = 14 and y - 2x = 1

(ii) y = x,y = 0 and 3x + 3y = 10

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