How do you write the equation for a circle with center at (2,3) that is tangent to the x-axis?

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How do you write the equation for a circle with center at (2,3) that is tangent to the x-axis?

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Answer 1 · 2018-07-26T19:02:20

${(x - 2)}^{2} + {(y - 3)}^{2} = 9$ $(x-2)^2+(y-3)^2=9$

Explanation:

Given that the center of circle is $(x_{1}, y_{1}) \equiv (2, 3)$ $(x_1, y_1)\equiv(2, 3)$ .

The above circle is tangent to the x-axis hence its radius

$r = y-coordinate = 3$ $r=\text{y-coordinate}=3$

hence the equation of the circle is given by following formula

${(x - x_{1})}_{1}^{2} + {(y - y_{1})}_{1}^{2} = r^{2}$ $(x-x_1)^2+(y-y_1)^2=r^2$

${(x - 2)}^{2} + {(y - 3)}^{2} = 3^{2}$ $(x-2)^2+(y-3)^2=3^2$

${(x - 2)}^{2} + {(y - 3)}^{2} = 9$ $(x-2)^2+(y-3)^2=9$

Answer 2 · 2018-07-26T19:55:44

${(x - 2)}^{2} + {(y - 3)}^{2} = 9$ $(x-2)^2+(y-3)^2=9$

Explanation:

$the equation of a circle in standard form is$ $"the equation of a circle in standard form is"$

$∙ x {(x - a)}^{2} + {(y - b)}^{2} = r^{2}$ $•color(white)(x)(x-a)^2+(y-b)^2=r^2$

$where (a, b) are the coordinates of the centre and r$ $"where "(a,b)" are the coordinates of the centre and r"$
$is the radius$ $"is the radius"$

$here (a, b) = (2, 3)$ $"here "(a,b)=(2,3)$

$r is the vertical distance from the x-axis to the centre$ $"r is the vertical distance from the x-axis to the centre"$

$r = y-coordinate of centre = 3$ $r="y-coordinate of centre "=3$

${(x - 2)}^{2} + {(y - 3)}^{2} = 9 \leftarrow equation of circle$ $(x-2)^2+(y-3)^2=9larrcolor(red)"equation of circle"$

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How do you write the equation for a circle with center at (2,3) that is tangent to the x-axis?

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