Find distance between intersections of circle and line?

02072019, 03:48 PM
(This post was last modified: 02072019 04:56 PM by kevin3g.)
Post: #1




Find distance between intersections of circle and line?
The equations are y=2x+3 and (x+1)^2+y^2=55. I know you can put them in solve and find x and y twice, but is there a faster way?


02072019, 06:05 PM
(This post was last modified: 02082019 01:08 PM by Albert Chan.)
Post: #2




RE: Find distance between intersections of circle and line?
Midpoint of intersecting points is closest to the circle center, (1,0), a line with slope 1/2
y = 2x + 3 y = 1/2*x 1/2 ; line passes circle center and midpoint > midpoint = (1.4, 0.2) Distance from midpoint to circle center = √(0.4² +0.2²) = √(20)/10 ~ 0.447 Distance from midpoint to intersecting points= √(55  20/100) = √(274/5) ~ 7.403 Distance between intersecting points ~ 2 * 7.403 ~ 14.8 Intersecting points = (1.4 ± t, 0.2 ± 2t) for some t t² + (2t)² = 5 t² = 274/5; matching distance square t = ±√(274)/5 > Intersecting points ~ (1.4 ± 3.31, 0.2 ± 6.62) = (4.71, 6.42) and (1.91, 6.82) 

02072019, 06:13 PM
(This post was last modified: 02072019 06:15 PM by kevin3g.)
Post: #3




RE: Find distance between intersections of circle and line?
Is there a way I can subtract/use the answers I get from the solve function easily without having to retype them?
{4.7106, 1.9106} {6.4212, 6.8212} 

02072019, 06:43 PM
(This post was last modified: 02072019 07:06 PM by ijabbott.)
Post: #4




RE: Find distance between intersections of circle and line?
(02072019 03:48 PM)kevin3g Wrote: The equations are y=2x+3 and (x+1)^2+y^2=55. I know you can put them in solve and find x and y twice, but is there a faster way? I don't know if it is faster, but: \[ y=2x+3, (x+1)^2+y^2=55 \\ (x+1)^2+(2x+3)^2=55 \\ x^2+2x+1+4x^2+12x+9=55 \\ 5x^2+14x45=0 \] If the quadratic equation \(ax^2+bx+c=0\) has real roots, the roots are separated by \(\frac{\sqrt{b^24ac}}{a}\), since \(\frac{b+\sqrt{b^24ac}}{2a}\frac{b\sqrt{b^24ac}}{2a} = \frac{2\sqrt{b^24ac}}{2a} = \frac{\sqrt{b^24ac}}{a}\). This gives the \(x\) separation of the intersections, \(\Delta x = \frac{\sqrt{b^24ac}}{a}\). The \(y\) separation, \(\Delta y\) will be twice that since \(y=2x+3\), so \(\Delta y = 2\Delta x\) (and the \(3\)s cancel). The distance between the intersection points \(D = \sqrt{(\Delta x)^2 + (\Delta y)^2} = \sqrt{(\Delta x)^2 + (2\Delta x)^2} = \sqrt{5(\Delta x)^2} = \sqrt{5\left(\frac{\sqrt{b^24ac}}{a}\right)^2} =\sqrt{5\left(\frac{b^24ac}{a^2}\right)} = \frac{\sqrt{5(b^24ac)}}{a}\). Plugging in the coefficients of the quadratic equation, \(D = \frac{\sqrt{5(14^2 (4 \cdot 5 \cdot (45)))}}{5} = \frac{\sqrt{5(196+900)}}{5} = \frac{\sqrt{5480}}{5} \approx 14.81\). EDIT: Corrected mistake in \(c\) spotted by Albert Chan. \(c\) should be \(45\), not \(41\). — Ian Abbott 

02082019, 12:33 AM
(This post was last modified: 02082019 02:31 PM by Albert Chan.)
Post: #5




RE: Find distance between intersections of circle and line?
Another way is to find angles of intersection.
Move the circle center to (0,0), line > y' = 2(x'  1) + 3 = 2x' + 1 Scale down to create a unit circle, line > y'' = 2 x'' + 1√55 > sin(z) = 2 cos(z) + 1√55 Use half angle formulas, t=tan(z/2), and let k=1√55: 2t/(1+t²) = 2 * (1t²)/(1+t²) + k 2t = 2  2t² + k + kt² (k2) t²  2t + (k+2) = 0 > t = 0.6605, 1.7328 > z = 66.89°, 239.98° Distance between intersecting point = 2 r sin(Δz/2) = 2 √(55) sin(173.09°/2) ~ 14.8 Intersecting hi point = (√(55) cos(66.89°)  1 , √(55) sin(66.89°)) ~ (1.91, 6.82) Intersecting lo point = (√(55)cos(239.98°)  1, √(55)sin(239.98°)) ~ (4.71, 6.42) Edit: this may be more accurate: sin(Δz/2) = Δt / √((1+t1²)(1+t2²)) 

02082019, 11:59 AM
Post: #6




RE: Find distance between intersections of circle and line?
You can see many examples, from ... http://www.hpprime.de/files/composite_f...ages_m.pdf


02082019, 12:08 PM
Post: #7




RE: Find distance between intersections of circle and line?
(02072019 03:48 PM)kevin3g Wrote: The equations are y=2x+3 and (x+1)^2+y^2=55. I know you can put them in solve and find x and y twice, but is there a faster way? Hey @kevin3g, you can solve the system between the two equations and once you found the intersection points coordinates you apply the usual formula (c=sqrt(((x1x2)^2)+((y1y2)^2))) to find their distance Honestly I dont know if there is a faster way than that given. Best, Aries 

02082019, 12:24 PM
Post: #8




RE: Find distance between intersections of circle and line?
More easy is ... https://www.wolframalpha.com/input/?i=circle+and+line


02082019, 12:45 PM
(This post was last modified: 02082019 01:44 PM by Albert Chan.)
Post: #9




RE: Find distance between intersections of circle and line?
Distance from point (x0, y0) to line Ax + By + C = 0 is abs(Ax0 + By0 + C)/√(A² + B²)
> Distance from circle center (1,0) to y=2x+3 is abs(2 + 0 + 3)/√5 = 1/√5 > Distance between intersecting points = 2 * √(55  1/5) = 2 √(274/5) ~ 14.8 

02082019, 01:05 PM
Post: #10




RE: Find distance between intersections of circle and line?
Continuing my previous post ...


02082019, 04:47 PM
Post: #11




RE: Find distance between intersections of circle and line?
(02082019 12:45 PM)Albert Chan Wrote: Distance from point (x0, y0) to line Ax + By + C = 0 is abs(Ax0 + By0 + C)/√(A² + B²) That's an elegant way to do it! — Ian Abbott 

09042019, 10:16 PM
(This post was last modified: 09102019 12:50 AM by Albert Chan.)
Post: #12




RE: Find distance between intersections of circle and line?
(02072019 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(xx0) + B(yy0) + (Ax0 + By0 + C) = 0 (yy0) = (A/B) (xx0)  (Ax0 + By0 + C)/B Matching pattern y' = mx' + c, then use h formula → \(\large \mathbf{ h= { A x_0 + B y_0 + C \over \sqrt{A^2 + B^2}}} \) 

09062019, 05:16 AM
(This post was last modified: 09062019 05:21 AM by teerasak.)
Post: #13




RE: Find distance between intersections of circle and line?
In CAS mode, use command
solve({y=2*x+3,(x+1)^2+y^2=55},{x,y}) then the calculator will return you 2 intersection points. Put that in a variable e.g m1 then use command distance(m1(1),m1(2)) you will get the distance between two points 

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