How do you write an equation with center (3,6), tangent to the x-axis?

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How do you write an equation with center (3,6), tangent to the x-axis?

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Answer 1 · 2016-07-18T08:41:14

The Question is not clear, but, I presume that what is reqd. is the eqn. of a circle having centre at (3,6) and the X-axis is tgt. to the reqd. circle.

$x^{2} + y^{2} - 6 x - 12 y + 9 + 0$ $x^2+y^2-6x-12y+9+0$

Explanation:

The Question is not clear, but, I presume that what is reqd. is the eqn. of a circle having centre at (3,6) and the X-axis is tgt. to the reqd. circle.

In that case, we know from Geometry that the $⊥$ $bot$ dist. from the centre to a tgt. of a circle is equal to the radius of the circle.

Hence, $r =$ $r=$ radius $= ⊥$ $=bot$ dist. from centre $(3, 6)$ $(3,6)$ to the X-axis
$=$ $=$ y-co-ord. of $C (3, 6) = 6$ $C(3,6)=6$ .

Accordingly, the eqn. of the circle is $: {(x - 3)}^{2} + {(y - 6)}^{2} = 6^{2}$ $: (x-3)^2+(y-6)^2=6^2$ , i.e.,

$x^{2} + y^{2} - 6 x - 12 y + 9 + 0$ $x^2+y^2-6x-12y+9+0$

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How do you write an equation with center (3,6), tangent to the x-axis?

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