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
please respond quickly
[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
please respond quickly[/quote
Need answers fast?
I will try to help.
Center of first circle 3,2
radius of circle is 6
Plot and id the two tangents.No cals required
Center of second circle 2,2
radius of circle is 4radical 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
Hello, wale!
Did you make any sketches?
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?
Find the coordinates of the endpoints of the diameter whose equation isCode:| | * * * * | * * * | o * | * * | * * | * * * | o (2,2) * * | * * - - - - + - - - - - - - - - - * * | * o | * * * | * | * * * |
We want the intersections of: .
Substitute [2] into [1]: .
. . . . . . .
Hence: .
Therefore, the endpoints of the diameter are: .