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write the standard form equation for the circle whose center is at (-2, 3) and that is tangent to the line 20x - 21y - 42 = 0.

a). x^2 - 4x + y^2 + 6y - 12 = 0
b). x^2 + 4x + y^2 - 6y - 12 = 0
c). x^2 + 4x + y^2 - 6y - 25 = 0

➷ Use the geometry rule:

(|(20(-2) - 21(3) - 42|) / square root of 20^2 + (-21)^2

Now you get:

(x + 2)^2 + (y - 3)^2 = 5^2

This can also be written as:

x^2 + 4x + y^2 - 6y - 12.

It would be option B

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for the friend request.

when a figure, such as triangle (xyz) is reflected over the line y = x to create x'y'z' the x-coordinates and y-coordinates of the original image (pre-image) will be reversed for the image. (e.g., (x, y) > (y, x)) this means that the pre-image position of the x- and y-coordinates switch places which in turn provides a mirror like image of the original xyz.  

i notice that when i draw a line segment from x to x', that the points are parallel from each other -- they are exactly across from each other and do not intersect with any other points.  

yes, think i would see the same characteristic if you drew the line segment connecting y with the reflecting line and then y' with the reflecting line because of the essence of reflections. because it is a mirror image, each point was effected the same way which leads me to conclude that they same would apply to y and y'.