question to with geometry of a circle

**cooltowns** · April 30th 2009, 10:15 AM

Hello i need help with the following question

I have attached my question

thanks

Kind Regards

**Opalg** · April 30th 2009, 12:30 PM

The centre of the circle must lie on the perpendicular bisector of PQ. This is the line through the midpoint (3,–1) of PQ, with gradient 2. So its equation is $y = 2x-7$ .

If a circle with centre at the point $(x,2x-7)$ touches the y-axis then its radius must be equal to x. If that circle passes through the point (1,0) then $(x-1)^2 + (2x-7)^2$ (the square of the distance from $(x,2x-7)$ to (1,0)) must therefore be equal to $x^2$ .

Solve the equation $(x-1)^2 + (2x-7)^2 = x^2$ to find the x-coordinate of the centre of the circle (two solutions, because it is a quadratic equation).

**stapel** · April 30th 2009, 01:54 PM

FYI for other viewers: The text of the exercise (from the graphic) is as follows:

Two circles touch the y-axis and pass through the points Q(1, 0) and P(5, -2). Find the equations of the two circles and determine their radii and centers.

**Soroban** · April 30th 2009, 07:11 PM

Hello, cooltowns!

Two circles touch the $y$ -axis and pass through the points $Q(1, 0)$ and $P(5, -2).$
Find the equations of the two circles and determine their radii and centers.

We want to locate the center of the circle $C(x,y)$
. . which is equidistant from $P, Q$ and the $y$ -axis.

Code:

        |
        |           C
      Y o  *  *  *  o (x,y)
        |         *  *
        |       *     *
        |   Q *        *
    - - + - o - - - - - * - - - -
        | (1,0)          o P
        |              (5,-2)

We want: . $CP \,=\,CQ \,=\,CY$

$\text{Then: }\;\underbrace{\sqrt{(x-5)^2 + (y-2)^2}}_{[1]} \;=\;\underbrace{\sqrt{(x-1)^2 + y^2}}_{[2]} \;=\;\underbrace{x}_{[3]}$

Square [2] and [3]: . $(x-1)^2 + y^2 \:=\:x^2 \quad\Rightarrow\quad x\:=\:\frac{y^2+1}{2}$ .(a)

Square [1] and [3]: . $(x-5)^2 + (y-2)^2 \:=\:x^2 \quad\Rightarrow\quad y^2 + 4y + 29 \:=\:10x$ .(b)

Substitute (a) into (b): . $y^2 + 4y + 29 \:=\:10\left(\frac{y^2+1}{2}\right) \quad\Rightarrow\quad y^2-y-6\:=\:0$

Then: . $(y-3)(y+2)\:=\:0 \quad\Rightarrow\quad y \:=\:3,\:\text{-}2 \quad\Rightarrow\quad x \:=\:5,\:\tfrac{5}{2}$

The circles are: . $\begin{Bmatrix}(x-5)^2 + (y-3)^2 \:=\:25 & C(5,3) & r = 5 \\ \\[-3mm]<br /> \left(x-\frac{5}{2}\right)^2 + (y+2)^2 \:=\:\frac{25}{4} & C\left(\frac{5}{2},\:\text{-}2\right) & r = \frac{5}{2} \end{Bmatrix}$

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