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It is easy to find the answer is either A or B, knowing the center coordinate $(2,2)$ and the value of $a$, which is $1$. However, I don't know how to calculate $b$ to find the equation.

Without finding $b$, though, I can still get the right answer A by plugging in the origin that is passed by the hyperbola.

But I still hope someone can teach me how to derive $b$.

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    $\begingroup$ hyperbolic-geometry is not the appropriate tag for your post. Do check it out though, it is a very interesting area. $\endgroup$
    – cosmo5
    Nov 29 '20 at 10:22
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Currently you have found $$\dfrac{(x-2)^2}{1^2}-\dfrac{(y-2)^2}{b^2}=1$$

Now use the condition that hyperbola passes through origin. Subbing $(x,y)=(0,0)$ in above will give value of $b$.

This is the standard way. Without using this condition, there is not enough information in the question to fix the hyperbola to be unique. There are infinitely many hyperbolas with vertices at the two points $(1,2)$ and $(3,2)$ only.

One needs minimum five points to uniquely determine a conic. Here we have $(0,0)$ and its reflections in the axes to fix our required hyperbola.

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