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:
\(\displaystyle (x- 3\sqrt{3})^2+ (y- 6)^2= 36\)
\(\displaystyle (x+ 3\sqrt{3})^2+ (y- 6)^2= 36\)