Prove or disprove that the point (1, sqrt3) lies on the circle that is centered at the origin and contains the point (0,2)

opatovaL 2021-03-08 Answered

Prove or disprove that the point (1, \(\sqrt{3}\)) lies on the circle that is centered at the origin and contains the point (0,2)

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Expert Answer

Mitchel Aguirre
Answered 2021-03-09 Author has 27246 answers

The standard equation of a circle with center (h,k) and radius r is given by:
\(\displaystyle{\left({x}−{h}\right)}^{{2}}+{\left({y}−{k}\right)}^{{2}}={r}^{{2}}\)
Using \((h,k)=(0,0)\) and \((x,y)=(0,2)\), we solve for r2:
\((0-0)2+(2-0)2=r2\)
\(4=r2\)
So, the equation of the circle is:
\(\displaystyle{x}^{{2}}+{y}^{{2}}={4}\)
The point (1,\(\sqrt{3}\)) lies on the circle if it satisfies the equation so we check:
\(\displaystyle{1}^{{2}}+{\left(√{3}\right)}^{{2}}=?{4}\)
\(1+3=?4\)
\(4=4\)
Since it satisfies the equation, then \(\displaystyle{\left({1},√{3}\right)}\) lies on the circle.

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