Circle problems and completing the square

[Kitobeirens]

Here is the question i cant do:

A circle passes through points Q(0,3) and R(0,9) and touches the x axis work out two possible equations

 

[soroban]

Hello, Kitobeirens!

I don't suppose you made a sketch . . .

A circle passes through points Q(0,3) and R(0,9) and touches the x-axis.
Work out two possible equations.

        |
        |     * * *
        | *           *
   (0,9)*               *
       *|                *
        |
      * |      (x,6)      *
      * |       *         *
      * |     * :         *
        |   *   :
       *| * 6   :6       *
   (0,3)*       :       *
        | *     :     *
    ----+-----*-*-*----------
        | - x - :
        |

We see that the center is at $\displaystyle y = 6.$

Hence, the radius is $\displaystyle 6.$

The distance from the center $\displaystyle (x,6)$ to $\displaystyle (0,3)$ must also be 6.

. . We have: .$\displaystyle \sqrt{(x-0)^2 + (6-3)^2} \:=\:6 \quad\Rightarrow\quad \sqrt{x^2 + 9} \:=\:6$

. . Square: .$\displaystyle x^2 + 9 \:=\:36 \quad\Rightarrow\quad x^2 \:=\:27 \quad\Rightarrow\quad x \:=\:\pm3\sqrt{3}$

The centers of the circles are: $\displaystyle \left(\pm3\sqrt{3},\,6\right)$. .The radius is 6.


The equations are: .$\displaystyle (x\pm3\sqrt{3})^2 + (y-6)^2 \:=\:36$

 

[HallsofIvy]

Geometry- the perpedicular bisector of any secant in a circle passes through the center of the circle. If (0, 3) and (0, 9) are points on the circle, then the line y= (3+ 9)/2= 6 passes through the center. That is, the center is the point (x, 6) for some x. The circle "touches the x-axis" rather than crossing it so the x- axis is a tangent line to the circle and so the line from (x, 6) to (x, 0) is a radius. That has length 6 so the radius of the circle is 6. Since that is true, the distance from (x, 6) to either of (0, 3) or (0, 9) must be 6. That is, $\displaystyle \sqrt{(x- 0)^2+ (6- 3)^2}= \sqrt{(x- 0)^2+ (6- 9)^2}= \sqrt{x^2+ 9}= 6$. Squaring both sides, $\displaystyle x^2+ 9= 36$, $\displaystyle x^2= 27$, $\displaystyle x= \pm\sqrt{27}= \pm3\sqrt{3}$. There are two possible centers having centers at $\displaystyle (3\sqrt{3}, 6)$ or $\displaystyle (-3\sqrt{3}, 6)$ and radius 6:
$\displaystyle (x- 3\sqrt{3})^2+ (y- 6)^2= 36$
$\displaystyle (x+ 3\sqrt{3})^2+ (y- 6)^2= 36$