The circle whose center is located in the first coordinate quarter touches the $ Ox $ axis at the $ M $ point, intersects two hyperbolas $ y_1 = \dfrac{k_1}{x} $ and $ y_2 = \dfrac{ k_2}{x} $ $ (k_1, k_2> 0) $ at the points $ A $ and $ B $ such that the line $ AB $ passes through the origin $ O $. It is known that $ \dfrac {4} {k_1} + \dfrac {1}{k_2} = 4 $. Find the smallest possible length of the $ OM $ segment. In response, write down the square of the length of the segment $ OM$.
I know if a circle equation is $(x - h)^2 + (y - k)^2 = a^2$ , when it touches x-axis then, $k = a$ and so the equation will be $(x - h)^2 + (y - a)^2 = a^2 => x^2 + y^2 - 2hx - 2ay + h^2 = 0$. The length of $OM$ will be $h$. The equation of $AB$ will be in $y = mx$ form.
Now, from $ y_1 = \dfrac{k_1}{x_1} $ and $ y_2 = \dfrac{ k_2}{x_2} $ $ (k_1, k_2> 0) $ , if we will $\dfrac{1}{y_1} + \dfrac{1}{y_2} = \dfrac{x_1}{k_1} + \dfrac{x_2}{k_2} = p $ and comparing it with $ \dfrac {4} {k_1} + \dfrac {1}{k_2} = 4 $ we get $x_1 = 4; x_2 = 1; p = 4 $.
I can't get the idea to proceed further. Please help me with the solution.
Consider $y=mx$ as the equation of the line $OA$ and find the coordinates of $A$ and $B$
$$B=\left\{\frac{k_1}{\sqrt{k_1 m}},\sqrt{k_1 m}\right\};A=\left\{\frac{k_2}{\sqrt{k_2 m}},\sqrt{k_2 m}\right\}$$
We have $$OA^2=k_2 m+\frac{k_2}{m};\;OB^2=k_1 m+\frac{k_1}{m}$$
Now apply the theorem which states that $OM^2=OB\cdot OA$
$$OM^2=f(m)=\sqrt{k_1 m+\frac{k_1}{m}} \sqrt{k_2 m+\frac{k_2}{m}}$$
and set derivative equal to zero to find extrema.
$$f'(m)=\frac{k_1 k_2 \left(m^4-1\right)}{m^3 \sqrt{k_1 \left(m+\frac{1}{m}\right)} \sqrt{k_2 \left(m+\frac{1}{m}\right)}}$$
$f'(m)=0\to m=1$ since $m=-1$ gives a line crossing the second and fourth quadrant.
Since $f'(m)>0$ for $m>1$ we can conclude that $m=1$ is a minimum
$$OM^2=2 \sqrt{k_1k_2}$$
knowing that $\frac{4}{k_1}+\frac{1}{k_2}=4$ we can conclude that the minimum for $OM^2$ is $$OM^2=\frac{k_1}{\sqrt{k_1-1}}$$
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