The point on the circle x^2+y^2=4 whose dista | vedclass

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(C) Let the point on the circle be $(h, k)$. Since it lies on the circle $x^2+y^2=4$,we have $h^2+k^2=4$. The distance of $(h, k)$ from the line $4x+3

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$(2,0)$

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(C) Let the point on the circle be $(h, k)$. Since it lies on the circle $x^2+y^2=4$,we have $h^2+k^2=4$. The distance of $(h, k)$ from the line $4x+3y-12=0$ is given by $\frac{|4h+3k-12|}{\sqrt{4^2+3^2}} = \frac{4}{5}$. This simplifies to $|4h+3k-12|=4$,which gives two cases: $4h+3k=16$ or $4h+3k=8$. Case $1$: $4h+3k=16 \Rightarrow k = \frac{16-4h}{3}$. Substituting into $h^2+k^2=4$ gives $h^2 + (\frac{16-4h}{3})^2 = 4$,which simplifies to $25h^2-128h+220=0$. The discriminant $D = 128^2 - 4(25)(220) = 16384 - 22000 Case $2$: $4h+3k=8 \Rightarrow k = \frac{8-4h}{3}$. Substituting into $h^2+k^2=4$ gives $h^2 + (\frac{8-4h}{3})^2 = 4$,which simplifies to $25h^2-64h+28=0$. Solving for $h$ using the quadratic formula: $h = \frac{64 \pm \sqrt{64^2 - 4(25)(28)}}{2(25)} = \frac{64 \pm \sqrt{4096 - 2800}}{50} = \frac{64 \pm \sqrt{1296}}{50} = \frac{64 \pm 36}{50}$. This gives $h = \frac{100}{50} = 2$ or $h = \frac{28}{50} = \frac{14}{25}$. If $h=2$,$k = \frac{8-4(2)}{3} = 0$. If $h=\frac{14}{25}$,$k = \frac{8-4(14/25)}{3} = \frac{200-56}{75} = \frac{144}{75} = \frac{48}{25}$. Thus,the points are $(2,0)$ and $(\frac{14}{25}, \frac{48}{25})$.

The point on the circle $x^2+y^2=4$ whose distance from the line $4x+3y-12=0$ is $4/5$ units is equal to

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