Question

Help

Answer 1

@jim_thompson5910

Answer 2

I would use implicit differentiation

x^2+y^2=r^2
d/dx[x^2+y^2]=d/dx[r^2]
2x + 2y*dy/dx = 0

now solve for dy/dx

2x + 2y*dy/dx = 0
2y*dy/dx = -2x
dy/dx = -2x/2y
dy/dx = -x/y


so the slope of any point on the circle (x,y) is described as `dy/dx = -x/y`

Answer 3

now turn to `ax+by+c=0`

what is the slope of this line?

Answer 4

y=-a/bx-c/b?

Answer 5

hopefully you see how the slope is `-a/b` ?

Answer 6

yea

Answer 7

slope of any tangent line to `x^2+y^2=r^2` is `-x/y`

slope of `ax+by+c=0` is `-a/b`

we want these slopes to be the same, so,

-a/b = -x/y
-ay = -bx
-ay+bx = 0

hmm I'm stuck at this part. Let me think.

Answer 8

hmm that's a clever way to do it, let me confirm if it works or not

Answer 9

yes, I'm able to construct it in geogebra. So it works

Answer 10

ok cool