Question
If one end of a diameter of the circle $x^{2}+y^{2}-10 x+6=0$ is at $(1,2)$, then the other end is at(a) $(9,-2)$(b) $(2,-9)$(c) $(5,0)$(d) $(9,0)$
Step 1
We can rewrite this equation in the standard form $(x-a)^{2}+(y-b)^{2}=r^{2}$, where $(a,b)$ is the center of the circle and $r$ is the radius. Show more…
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