The line 3x - 2y = k meets the circle {x^2} + | vedclass

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(B) The equation of the line is $3x - 2y = k$,which can be rewritten as $y = \frac{3}{2}x - \frac{k}{2}$.Comparing this with $y = mx + c$,we get $m =

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$52{r^2}$

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(B) The equation of the line is $3x - 2y = k$,which can be rewritten as $y = \frac{3}{2}x - \frac{k}{2}$.Comparing this with $y = mx + c$,we get $m = \frac{3}{2}$ and $c = -\frac{k}{2}$.The circle is ${x^2} + {y^2} = 4{r^2}$,which has a radius $a = 2r$.$A$ line $y = mx + c$ is tangent to the circle ${x^2} + {y^2} = a^2$ if $c^2 = a^2(1 + m^2)$.Substituting the values,we get $\left(-\frac{k}{2}\right)^2 = (2r)^2 \left(1 + \left(\frac{3}{2}\right)^2\right)$.$\frac{k^2}{4} = 4{r^2} \left(1 + \frac{9}{4}\right)$.$\frac{k^2}{4} = 4{r^2} \left(\frac{13}{4}\right)$.$\frac{k^2}{4} = 13{r^2}$.Therefore,${k^2} = 52{r^2}$.

The line $3x - 2y = k$ meets the circle ${x^2} + {y^2} = 4{r^2}$ at only one point,if ${k^2} =$

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