# coordinate geometry of circles

• Apr 5th 2010, 12:28 PM
wale
coordinate geometry of circles
1. (x-3)^2 +(y-2)^2=36
find equations of the two horizontal tangents to the circle

2.(x-2)^2 +(y-2)^2 =32
Find coordinates of the endpoints of the diameter whose equation is y=x

• Apr 5th 2010, 01:21 PM
bjhopper
[quote=wale;487298]1. (x-3)^2 +(y-2)^2=36
find equations of the two horizontal tangents to the circle

2.(x-2)^2 +(y-2)^2 =32
Find coordinates of the endpoints of the diameter whose equation is y=x

I will try to help.
Center of first circle 3,2

Plot and id the two tangents.No cals required

Center of second circle 2,2
Plot the center point and draw in y=x start at origin line has slope = 1 extend line above and below the x axis
draw in two slope diagrams using the center as reference point but the rise and the run must be 4 so that the hypothenus will be 4 rad 2. No cals required

bjh
• Apr 5th 2010, 02:29 PM
Soroban
Hello, wale!

Did you make any sketches?

Quote:

$1.\;\;(x-3)^2 +(y-2)^2\:=\:36$

Find the equations of the two horizontal tangents to the circle.

Code:

              |               | (3,8)         - - - + * o * - - -             * |  :    *           *  |  :      *         *    |  : r=6    *               |  :         *    |  :        *         *    |  o(3,2)    *         *    |  :        *               |  :       ---*----+---:--------*-----           *  |  :      *             * |  :    *         - - - + * o *  - - -               | (3,-4)               |

Got it?

Quote:

$2.\;\;(x-2)^2 +(y-2)^2 \:=\:32$

Find the coordinates of the endpoints of the diameter whose equation is $y=x$

Code:

              |               | * * *             * |        *  *           *  |          o         *    |        *  *               |      *         *    |    *      *         *    |  o (2,2)  *         *    | *          *       - - - - + - - - - - - - - - -         *  * |            *           o  |          *         *  * |        *               | * * *               |

We want the intersections of: . $\begin{Bmatrix}[1] & (x-2)^2 + (y-2)^2 \:=\:32 \\ [2] & x\:=\:y \end{Bmatrix}$

Substitute [2] into [1]: . $(x-2)^2 + (x-2)^2 \:=\:32 \quad\Rightarrow\quad 2(x-2)^2 \:=\:32$

. . . . . . . $(x-2)^2 \:=\:16 \quad\Rightarrow\quad x-2 \:=\:\pm4 \quad\Rightarrow\quad x \:=\:2 \pm 4$

Hence: . $x \;=\;6,\:-2 \quad\Rightarrow\quad y \;=\;6,\:-2$

Therefore, the endpoints of the diameter are: . $(6,6)\,\text{ and }\,(\text{-}2,\text{-}2)$