how to find the equation of a circle tangent to the y-axis given two points A(1;0), B(5;0)?

Question

So I have a circle that passes by A(1;0) and B(5;0) and is tangent to the y-axis. I need to find the equation of the circle.

Answer 1

If the equation of the circle is $(x-a)^2+(y-b)^2=r^2$

$$(1-a)^2+(0-b)^2=r^2=(5-a)^2+(0-b)^2$$

$$\implies a=3$$

Put $x=0$ in $$(x-3)^2+(y-b)^2=r^2$$

$$(0-3)^2+(y-b)^2=r^2\iff(y-b)^2=r^2-9$$

For tangency, $r^2-9=0$

Put the values of $a,r^2$ in $$(1-a)^2+(0-b)^2=r^2$$ to find $b$

Answer 2

The center of circle in on the line $x=3$, this is bisector of line on x axis between points (1, 0) and (5, 0). The circle is tangent to y axis so the radius is $r = 3 ⇒ x_c=3$. The center have distance $y_c$ from x axis :

$y_c=\sqrt {3^2 - 2^2}= \sqrt 5$

Where $x_c$ and $y_c$ are the coordinates of center; so we have:

$(x-3)^2 + (y-\sqrt 5)^2=3^2 =9$

Answer 3

The centre $O$ lies on the perpendicular bisector of $AB$. $O$ lies on vertical line passing through $(3,0)$ the midpoint of $AB$. Let $O(3,Y)$. Let $C$ be the point of contact of tangent. $OC\bot $ y axis so $OC=r=3$. $$AO=3\\ \sqrt{(3-1)^2+(Y-0)^2}=3\\ Y=\pm \sqrt5$$ The coordinates of $O$ are $(3,\pm \sqrt5)$ So the equation of the circle is given by $(x-3)^2+(y\mp \sqrt5)^2=9$

Note: there are two such circles.