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Question

# Write an equation for the circle that satisfies each set of conditions.center at $(-1,2)$, tangent to $x$-axis

Solution

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Given the center at $(-1,2)$ and tangent to $x$-axis, we can conclude that one endpoint is at $(-1,0)$.

Solve for $r$ using distance formula.

\begin{aligned} d&=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}\\ &=\sqrt{((-1)-(-1))^2+(0-2)^2}\\ &=\sqrt{(0)^2+(-2)^2}\\ &=\sqrt{4}\\ &=2 \end{aligned}

Hence $r=2$.

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