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(C) The equation of the circle with centre $(2,0)$ and radius $r$ is $(x-2)^2 + y^2 = r^2$.The equation of the ellipse is $x^2 + 4y^2 = 36$,which impl

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$92$

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(C) The equation of the circle with centre $(2,0)$ and radius $r$ is $(x-2)^2 + y^2 = r^2$.The equation of the ellipse is $x^2 + 4y^2 = 36$,which implies $y^2 = \frac{36-x^2}{4}$.Substituting $y^2$ into the circle equation:$(x-2)^2 + \frac{36-x^2}{4} = r^2$$4(x^2 - 4x + 4) + 36 - x^2 = 4r^2$$4x^2 - 16x + 16 + 36 - x^2 = 4r^2$$3x^2 - 16x + 52 - 4r^2 = 0$.For the circle to be inscribed,the quadratic equation must have a discriminant $D = 0$ for tangency:$D = (-16)^2 - 4(3)(52 - 4r^2) = 0$$256 - 12(52 - 4r^2) = 0$$256 - 624 + 48r^2 = 0$$48r^2 = 368$$12r^2 = \frac{368}{4} = 92$.

If the radius of the largest circle with centre $(2,0)$ inscribed in the ellipse $x^2+4y^2=36$ is $r$,then $12r^2$ is equal to

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