Let a circle C _ 1 be obtained on rolling the circle x ^ 2 + y ^ 2...

We have,

${x}^{2} + {y}^{2} - 4x - 6y + 11 = 0$

$⇒ {\left(x - 2\right)}^{2} + {\left(y - 3\right)}^{2} = 2$

Centre$≡ \left(2, 3\right)$

Radius $r = \sqrt{2}$ units

Equation of tangent at \left(3, 2\right)$is</p><p>$3x + 2y - 2\left(x + 3\right) - 3\left(y + 2\right) + 11 = 0$</p><p>$⇒ x = y = 1 . . . \left(1\right)$</p><p>On rolling the given circle upward$4$units on the tangent \left(1\right)$, centre of the circle also moving $4$ unit upward, so

Centre of ${C}_{1}$ is

$= \left(2 + 2\sqrt{2}, 3 + 2\sqrt{2}\right)</p><p>A\left(2 + 2\sqrt{2}, 3 + 2\sqrt{2}\right)</p><p>Centre$ of {C}_{2}$is</p><p>$B\left(4 + 2\sqrt{2}, 1 + 2\sqrt{2}\right)$</p><p>So,$M\left(2 + 2\sqrt{2}, 0\right)$and$N\left(4 + 2\sqrt{2}, 0\right)$.</p><p>$∴$area of trapezium</p><p>$= \dfrac{1}{2}\left(AM + BN\right) \times MN$</p><p>$= \dfrac{1}{2}\left(4 + 4\sqrt{2}\right)2 = 4\left(1 + \sqrt{2}\right) sq. units