# Thread: question to with geometry of a circle

1. ## question to with geometry of a circle

Hello i need help with the following question

I have attached my question

thanks

Kind Regards

2. 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 $\displaystyle y = 2x-7$.

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

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

3. 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.

4. Hello, cooltowns!

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

We want to locate the center of the circle $\displaystyle C(x,y)$
. . which is equidistant from $\displaystyle P, Q$ and the $\displaystyle y$-axis.
Code:
        |
|           C
Y o  *  *  *  o (x,y)
|         *  *
|       *     *
|   Q *        *
- - + - o - - - - - * - - - -
| (1,0)          o P
|              (5,-2)

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

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

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

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

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

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

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