Proof of Intersection of a Line and a Circle

Home , Intersection

P2(x2,y2)

P1(x1,y1) r C(xc,yc)

This problem is most easily solved if the circle is in implicit form:

2 2 2 (x − xc) + (y − yc) − r = 0 and the line is parametric:

x = x0 + ft

y = y0 + gt

Substituting for (parametric line) x and y into the circle equation gives a quadratic equation in t:

• Two roots of which give points on the line where cuts the circle.

p f(x − x ) + g(y − y ) ± r2(f 2 + g2) − (f(y − y ) − g(x − x ))2 t = c 0 c 0 c 0 c 0 (f 2 + g2)

• The roots maybe coincident if the line is tangential to the circle.

P1(x1,y1)

1 r C(xc,yc)

• If roots are imaginary then there is no intersection.

1 Proof:

The circle is in implicit form:

2 2 2 (x − xc) + (y − yc) − r = 0 (1) and the line is parametric:

x = x0 + ft

y = y0 + gt (2)

Substituting for (parametric line) x and y from Eqn. 2 into the circle equation, Eqn. 1, gives a quadratic equation in t gives:

2 2 2 (x0 + ft − xc) + (y0 + gt − yc) − r = 0 (3)

Let

xd = x0 − xc

yd = y0 − yc (4)

Then we may write Eqn. 3

2 2 2 (xd + ft) + (yd + gt) − r = 0 (5)

Expanding the quadratic parts in Eqn. 5 gives:

2 2 2 2 2 2 2 xd + 2fxdt + f t + yd + 2gydt + g t − r = 0 (6)

Express Eqn. 6 as a quadratic in t, so gather terms in t in Eqn. 6:

2 2 2 2 2 2 (f + g )t + (2fxd + 2gyd)t + xd + yd − r = 0 2 2 2 2 2 2 (f + g )t + 2(fxd + gyd)t + xd + yd − r = 0 (7)

2 2 Now the general solution of a quadratic of the form ax + bx + c = 0, gives the roots, xi, i = 1, 2 as:

√ −b ± b2 − 4ac x = (8) i 2a

So in our problem (a quadratic in t) from Eqn. 7 we can see that for Eqn. 8:

a = (f 2 + g2)

b = 2(fxd + gyd) 2 2 2 c = xd + yd − r

Substituting for a, b and c in Eqn. 8 we get roots of t, ti:

q 2 2 2 2 2 2 −2(fxd + gyd) ± (2(fxd + gyd)) − 4(xd + yd − r )(f + g ) t = i 2(f 2 + g2) q 2 2 2 2 2 2 −2(xd + gyd) ± 4(fxd + gyd) − 4(xd + yd − r )(f + g ) t = i 2(f 2 + g2) (9)

The expression of the right hand side of Eqn. 9 can be simpliﬁed in a few ways.

√ Firstly we can factor 4 from the term and the division by 2 then simpliﬁes the equation to:

q 2 2 2 2 2 2 −2(xd + gyd) ± 2 (fxd + gyd) − (xd + yd − r )(f + g ) t = i 2(f 2 + g2) q 2 2 2 2 2 2 −(fxd + gyd) ± (fxd + gyd) − (xd + yd − r )(f + g ) t = i (f 2 + g2) (10)

3 √ Let us now consider the term and simplify this, let:

2 2 2 2 2 2 S = (fxd + gyd) − (xd + yd − r )(f + g )) 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 S = f xd + 2fgxdyd + g yd − f xd − f yd + f r − g xd − g yd + g r (11)

2 2 2 2 Eqn 11 can be simpliﬁed as follows, some terms cancel out f xd, g yd and others can be gathered together r2, f, and g terms:

2 2 2 2 2 2 2 S = r (f + g ) − (f yd − 2fgxdyd + g xd 2 2 2 2 S = r (f + g ) − (fyd − gxd) 2 2 2 2 S = r (f + g ) − (−gxd + fyd) (12)

Substituting for xd and yd from Eqn. 4 in Eqn 12 we get:

2 2 2 2 S = r (f + g ) − (−g(x0 − xc) + f(y0 − yc)) 2 2 2 2 S = r (f + g ) − (g(xc − x0) − f(yc − y0))

√ Substituting for S from Eqn. 13 and xd and yd from Eqn. 4 in the term in Eqn. 10 gives us the solution:

p f(x − x ) + g(y − y ) ± r2(f 2 + g2) − (f(y − y ) − g(x − x ))2 t = c 0 c 0 c 0 c 0 (f 2 + g2)

Q.E.D