Question

Find the equation of a circle with center $(2,0)$ and passing through point $(3,\sqrt{3})$.

• A. $(x-2{)}^2+y^2=4$
• B. $(x+2{)}^2+y^2=4$
• C. $(x-2{)}^2+y^2=16$
• D. $(x+2{)}^2+y^2=16$

Answer 1

The correct option is A $(x-2{)}^2+y^2=4$

Let the radius be '$r$'.

Then the equation of the circle will be

$(x-2{)}^2+y^2=r^2$

Now, substituting the point $(3,\sqrt{3})$ in the above equation gives us $(3-2{)}^2+(\sqrt{3}{)}^2=r^2$

$\Rightarrow 1+3^2=r^2$ or $r^2=4$

Hence, the equation of the circle is $(x-2{)}^2+y^2=4$.