A circle C_1 passes through the origin O and has diameter 4 on the positive x-axis.

Question

A circle $C_1$ passes through the origin $O$ and has diameter $4$ on the positive $x$-axis. The line $y=2x$ gives a chord $OA$ of circle $C_1$. Let $C_2$ be the circle with $OA$ as a diameter. If the tangent to $C_2$ at the point $A$ meets the $x$-axis at $P$ and $y$-axis at $Q$, then $QA:AP$ is equal to:

Accepted Answer

$C_1:x^2+y^2-4x=0$. Intersect with $y=2x$: $5x^2-4x=0\Rightarrow x=0,\frac45$, so $A=\left(\frac45,\frac85\right)$.

$C_2$ has diameter $OA$, so tangent at $A$ is perpendicular to $OA$ (slope $2$), giving slope $-\frac12$:
$$y-\frac85=-\frac12\left(x-\frac45\right)\ \Longrightarrow\ x+2y=4.$$

$P=(4,0)$, $Q=(0,2)$. $QA=\left|A-Q\right|$, $AP=\left|P-A\right|$; by the section ratio along the line, $QA:AP=1:4$.

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