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Question

# Write an equation for the circle that satisfies the set of conditions.center at $(4,2)$, tangent to $x$-axis

Solution

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Given the center $(4,2)$ and knowing that the circle is tangent to $x$-axis, we can conclude that the radius is $2$.

Write in standard form using the center $(4,2)$ and $r=2$.

\begin{aligned} (x-h)^2+(y-k)^2&=r^2\\ (x-4)^2+(y-2)^2&=(2)^2&&\text{substitute \& simplify}\\ (x-4)^2+(y-2)^2&=4 \end{aligned}

Therefore the equation is $(x-4)^2+(y-2)^2=4$

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