Find distance between intersections of circle and line?

[Albert Chan]

(02-07-2019 03:48 PM)kevin3g Wrote:  The equations are y=2x+3 and (x+1)^2+y^2=55.

I just thought of a way, even if we do not know the distance for point to line formula.
At the end, I accidentally proved the "official" distance formula

First, shift the coordinate so the point (in this case, center of circle), is the origin.

y = 2x + 3 = 2(x+1) + 1 = 2x' + 1

With this shifted coordinates, find where the line hit the axis.
In other words, find points on line, P=(0, y0), Q=(x0, 0).

Let h = distance of origin to the line.

Area ΔOPQ = ½ |x0 y0| = ½ |PQ| h

\(\large \mathbf{h = {|x_0 y_0| \over \sqrt{x_0^2 + y_0^2 }}}\)

But, we can do better !
Assume the line has form y = m x + c, we have (x0, y0) = (-c/m, c)

h = | -c²/m | / √(c²/m² + c²)

\(\large \mathbf{h = {|c| \over \sqrt{m² + 1}}}\)

For this example, h = |1| / √(2² + 1) = 1/√5

Distance of chord = \(\large \mathbf{2 \sqrt{r^2 - \frac{c^2}{m^2+1}}}\) = 2 √(55 - 1/5) ≈ 14.8

For prove of official distance formula, rewrite Ax + By + C = 0

A(x-x0) + B(y-y0) + (Ax0 + By0 + C) = 0

(y-y0) = (-A/B) (x-x0) - (Ax0 + By0 + C)/B

Matching pattern y' = mx' + c, then use h formula → \(\large \mathbf{ h= { |A x_0 + B y_0 + C| \over \sqrt{A^2 + B^2}}} \)