How do you write an equation for a circle tangent to the y-axis and with the center (5, 6)?

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Answer 1 · 2016-09-01T05:02:07

$x^2+y^2-10x-12y+36=0$

The reqd. circle touches the Y-axis [ eqn. $x=0$ ] and has the Centre

at $C(5,6)$ .

From Geometry , we know that, in the given situation, the $bot$ -

distance from the Centre to the tgt. must equal the

Radius $r$ of the circle.

$:. r=|5|=5$ .

With Centre $(5,6) and r=5$ , the eqn. of circle is,

$(x-5)^2+(y-6)^2=5^2-12y+36=0$ . i.e.,

graph{x^2+y^2-10x-12y+36=0 [-22.8, 22.8, -11.4, 11.42]} $x^2+y^2-10x-12y+36=0$