A circle C_{1} passes through the origin O an | vedclass

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(A) The circle $C_{1}$ has its center at $(2, 0)$ and radius $2$. Its equation is $(x-2)^2 + y^2 = 4$,which simplifies to $x^2 + y^2 - 4x = 0$.Given t

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$1:4$

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(A) The circle $C_{1}$ has its center at $(2, 0)$ and radius $2$. Its equation is $(x-2)^2 + y^2 = 4$,which simplifies to $x^2 + y^2 - 4x = 0$.Given the line $y = 2x$,we substitute this into the circle equation: $x^2 + (2x)^2 - 4x = 0$,so $5x^2 - 4x = 0$. Thus,$x = 0$ (at $O$) or $x = 4/5$ (at $A$).For $x = 4/5$,$y = 2(4/5) = 8/5$. So,$A = (4/5, 8/5)$.The slope of $OA$ is $m = 2$. Let $	heta$ be the angle $OA$ makes with the $x$-axis,so $	an 	heta = 2$.The tangent to $C_{2}$ at $A$ is perpendicular to the radius $OA$. Let the tangent line be $L$. Since $OA$ has slope $2$,the tangent $L$ has slope $-1/2$.The equation of the tangent at $A(4/5, 8/5)$ is $y - 8/5 = -1/2(x - 4/5)$,which simplifies to $y - 8/5 = -1/2x + 2/5$,or $x + 2y = 4$.The $x$-intercept $P$ is found by setting $y=0$,giving $x=4$,so $P = (4, 0)$.The $y$-intercept $Q$ is found by setting $x=0$,giving $2y=4$,so $y=2$,$Q = (0, 2)$.The distance $AP = \sqrt{(4 - 4/5)^2 + (0 - 8/5)^2} = \sqrt{(16/5)^2 + (-8/5)^2} = \sqrt{256/25 + 64/25} = \sqrt{320/25} = \frac{8\sqrt{5}}{5}$.The distance $QA = \sqrt{(0 - 4/5)^2 + (2 - 8/5)^2} = \sqrt{(-4/5)^2 + (2/5)^2} = \sqrt{16/25 + 4/25} = \sqrt{20/25} = \frac{2\sqrt{5}}{5}$.The ratio $QA : AP = \frac{2\sqrt{5}}{5} : \frac{8\sqrt{5}}{5} = 2 : 8 = 1 : 4$.

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