This is a question I found: Given two circles intersecting orthogonally having the length of common chord 24/5 units the radius of one of the circles is 3 units then what is the sine of the angle between the direct common tangents? The answer given is 4√6/25. How do I approach this problem? I know that direct common tangents meet at a point that externally divides line joining centres of circles in ratio of their radii, but I am getting negative values when I try to solve this. How do I do this?
[Math] Sine of angle between direct common tangents of two circles
circlesconic sectionsgeometry
Related Solutions
Here's an argument that could probably be streamlined a bit ...
Restating the problem:
Let $\bigcirc A$, $\bigcirc B$, $\bigcirc C$ ---of radii $a$, $b$, $c$--- be pairwise externally tangent, with points of tangency $D$, $E$, $F$ as shown in the image. Let the line tangent to $\bigcirc A$ and $\bigcirc B$ at $F$ meet $\bigcirc C$ in chord $\overline{PQ}$ with midpoint $M$; define $p := |\overline{PM}| = |\overline{QM}|$. Then
$$p = \frac{2c\sqrt{ab}}{a+b} \tag{$\star$}$$
Extend $\overline{FE}$ and $\overline{FD}$ to meet $\bigcirc C$ again at $A^\prime$ and $B^\prime$. Note that $\triangle AEF$ and $\triangle CEA^\prime$ are isosceles, with congruent base angles at $E$ and therefore also at $F$ and $A^\prime$; thus, $\overline{A^\prime C}\parallel\overline{AF}$. Likewise, $\overline{B^\prime C}\parallel\overline{BF}$, making $A^\prime$, $B^\prime$, $C$ collinear on a diameter of $\bigcirc C$.
By Thales' Theorem, $\angle A^\prime PB^\prime$ is a right angle; because $\overline{PM}\perp\overline{A^\prime B^\prime}$, we have the classical construction of a geometric mean: $$|\overline{PM}|^2 = |\overline{A^\prime M}||\overline{B^\prime M}| \tag{1}$$
(This is why I chose $p$ to name the half-chord, instead of the full length of $\overline{PQ}$.) Now our task reduces to a matter of rewriting the right-hand side.
Observe that the perpendiculars at $D$, $E$, $F$ necessarily concur at a point equidistant from them; that is, they meet at center $I$ of the incircle (of radius $r$) of $\triangle ABC$. Of course, $\overline{AI}$ and $\overline{BI}$ are angle bisectors.
Since $\triangle A^\prime M F\sim \triangle IFA$ and $\triangle B^\prime MF\sim \triangle IFB$, we have $$\begin{align} \frac{|\overline{A^\prime M}|}{|\overline{MF}|} = \frac{r}{a} \quad\text{and}\quad \frac{|\overline{B^\prime M}|}{|\overline{MF}|} = \frac{r}{b} &\quad\implies\quad \frac{|\overline{B^\prime M}|}{|\overline{A^\prime M|}} = \frac{a}{b} \tag{2} \\ &\quad\implies\quad |\overline{A^\prime M}| = 2c\;\frac{b}{a+b} \quad\text{and}\quad |\overline{B^\prime M}| = 2c\;\frac{a}{a+b} \end{align}$$
(The above improves on my first pass at this proof. (See the revision history.) There may yet be a better argument that incorporates this proportionality relation: $$\frac{|\overline{A^\prime M}|}{b} = \frac{|\overline{A^\prime B^\prime}|}{|\overline{AB}|} = \frac{|\overline{B^\prime M}|}{a} \tag{2.5}$$ Be that as it may ...) Thus, $(1)$ becomes $$p^2 = \frac{4abc^2}{(a+b)^2} \qquad\to\qquad p = \frac{2c\sqrt{ab}}{a+b} \tag{3}$$ as claimed. $\square$
For the original question, we take $a = 4$, $b = 16$, $c = D/2$, so that the length of the chord is $$2p = \frac{4 \cdot D/2 \cdot \sqrt{4\cdot 16}}{4+16} = \frac{4D}{5}$$
Get the angle between the imaginary inner common tangents when two ellipse intersect with each other
New updated code.
This seems to work better:
'''
angle between inner tangents between two intersecting ellipses, newer version
'''
import numpy as np
import math
import cmath
'''
####### FUNCTIONS IMPORTANT FOR THE CALCULATION #######
'''
def homogenize(x):
return np.array([x[0], x[1], 1])
# transforming the vector of six coefficients
# of a quadratic equation with two variables into a matrix of
# the corresponding conic
def conic_from_equation(eq):
'''
eq is an np.array vector of the quadratic equation's six coefficients
eq[0]*x**2 + eq[1]*x*y + eq[2]*y**2 + eq[3]*x + eq[4]*y + eq[5] = 0
conc is a symmetric matrix of the bilinier form of the quadratic equation
'''
return np.array([[2*eq[0], eq[1], eq[3]],
[ eq[1], 2*eq[2], eq[4]],
[ eq[3], eq[4], 2*eq[5]]]) / 2
# transformaing the matrix of a conic into a vector of six coefficients
# of a quadratic equation with two variables
def equation_from_conic(C):
'''
c[0]*x**2 + c[1]*x*y + c[2]*y**2 + c[3]*x + c[4]*y + c[5] = 0
'''
return np.array([C[0,0], 2*C[0,1], C[1,1], 2*C[0,2], 2*C[1,2], C[2,2]])
# obtaining the conic dual to a given conic,
# both represented as symmetric 3x3 matrices
def dual_conic(conic):
return np.linalg.inv(conic)
# given a pair of conics, as a pair of symmetric 3x3 matrices,
# calculate the vector k = (k[0], k[1], k[2]) of values for each of which
# the conic c1 - k[i]*c2 from the pencil of conics c1 - t*c2
# is a degenerate conic (the anti-symmetric product of a pair of linear forms)
# and also find the matrix U
# of the projective transformation that simplifies the geometry of
# the pair of conics, the geometry of the pencil c1 - t*c2 in general,
# as well as the geometry of the three degenerate conics in particular
def trivialization_of(conic1, conic2):
'''
c1 and c2 are 3 by 3 symmetric matrices of two conics (possibly dual)
'''
c21 = np.linalg.inv(conic2).dot(conic1)
k, U = np.linalg.eig(c21)
return k, U
# find the common points, i.e. points of intersection of a pair of conics
# represented by a pair of symmetric matrices
def find_intersection_points(c1, c2):
k, U = trivialization_of(c1, c2)
L1 = (U.T).dot((c1 - k[0]*c2).dot(U))
L2 = (U.T).dot((c1 - k[1]*c2).dot(U))
x_0 = cmath.sqrt(-(L2[2,2] / L2[0,0]))
y_0 = cmath.sqrt(-(L1[2,2] / L1[1,1]))
sol = np.array([[ x_0, y_0, 1],
[-x_0, -y_0, 1],
[-x_0, y_0, 1],
[ x_0, -y_0, 1]])
sol = sol.dot(U.T)
return sol
# find the three coefficients of all four common tangents to a pair of conics
# make sure that the conics do have four common tangents,
# which is the generic situation
def find_common_tangents(c1, c2):
dc1 = dual_conic(c1)
dc2 = dual_conic(c2)
return find_intersection_points(dc1, dc2)
# if the two conics are ellipses, they do not intersect the line at infinity.
# by polarity, the poles of the line at infinity are
# the (affine and in fact metric) centers of the ellipses and are located
# inside the ellipses each. We use them to detect the inner tangents and
# to orient the vectors normal to these tangents in a way
# so that the angle between these two normal vectors is the vector on the picture
def dual_to_infinity_line(conic):
p = dual_conic(conic).dot(np.array([0,0,1]))
return homogenize( p[0:2] / p[2] )
# find the common inner tangents between a pair of conics,
# represented as 3x3 symmetric matrices, with properly oriented normal vectors
def find_common_inner_tangents(c1, c2):
tn = find_common_tangents(c1, c2)
p1 = dual_to_infinity_line(c1)
p2 = dual_to_infinity_line(c2)
# boolean vector, true if inner tangent false if not
is_inner_tangent = np.logical_or(tn.real.dot(p1)*tn.real.dot(p2) < 0, tn.imag.dot(p1)*tn.imag.dot(p2) < 0)
tn = tn[ is_inner_tangent, : ]
# reorient one of the normal vectors so that they give us the proper angle
if tn.real[0,:].dot(p1)*tn.real[1,:].dot(p1) > 0 or tn.imag[0,:].dot(p1)*tn.imag[1,:].dot(p1) > 0:
tn[1,:] = - tn[1,:]
return tn
# calculate the angle between a pair of line, given by their three coefficients
def angle_bn_C_lines(line1, line2):
theta = line1[0:2].dot(line2[0:2]) / ( cmath.sqrt( line1[0:2].dot(line1[0:2])*line2[0:2].dot(line2[0:2])) )
return cmath.acos(theta).conjugate()
# calculate the proper angle between the inner or complex tangents of two quadratic equations
# that represent a pair of conics, hopefully non-intersecting
def angle_bn_inner_tangents_of(eq_of_conic1, eq_of_conic2):
c1 = conic_from_equation(eq_of_conic1)
c2 = conic_from_equation(eq_of_conic2)
tn = find_common_inner_tangents(c1, c2)
return angle_bn_C_lines(tn[0,:], tn[1,:])
'''
####### END OF FUCTIONS IMPORTANT FOR THE CALCULATION #######
'''
'''
####### auxiliary functions to handle the test scenario:
'''
def cos_sin(angle_deg):
return math.cos(angle_deg*math.pi/180), math.sin(angle_deg*math.pi/180)
def rotation(cs_sn):
return np.array([[cs_sn[0], -cs_sn[1]],
[cs_sn[1], cs_sn[0]]])
def isom(angle, translation):
'''
isometry from global coordinate system (world system)
to conic-aligned coordinate system (conic attached)
'''
cos_, sin_ = cos_sin(-angle)
tr = - rotation((cos_, sin_)).dot(translation)
return np.array([[ cos_, -sin_, tr[0]],
[ sin_, cos_, tr[1]],
[ 0, 0, 1 ]])
# calculating the conic defined by a pair of axes' lengths,
# axes rotation angle and center of the conic
def Conic(major, minor, angle, center):
D = np.array([[minor**2, 0, 0],
[ 0, major**2, 0],
[ 0, 0, -(minor*major)**2]])
U = isom(angle, center)
return (U.T).dot(D.dot(U))
'''
####### end of auxiliary functions
'''
'''
test
'''
a1 = 5
b1 = 4.95
cntr1 = np.array([0,0])
w1 = 45
Eq1 = equation_from_conic(Conic(a1, b1, w1, cntr1))
Co1 = conic_from_equation(Eq1)
a2 = 1
b2 = .95
cntr2 = np.array([3,0])
w2 = 120
Eq2 = equation_from_conic(Conic(a2, b2, w2, cntr2))
Co2 = conic_from_equation(Eq2)
t4 = find_common_tangents(Co1, Co2)
print('all four tangents: ')
print(t4)
print('')
it = find_common_inner_tangents(Co1, Co2)
print('the inner tangents only: ')
print(it)
print('')
print('angle between inner tangents calculated by algorithm: ')
print( angle_bn_inner_tangents_of(Eq1, Eq2) )
print('')
print('angle calculation for circular approximation of the test case: ')
print('angle of internal tangents of a circle: ')
print(2*cmath.asin((a1+a2)/cntr2[0]))
print('angle of external tangents of a circle: ')
print(2*cmath.asin(abs(a1-a2)/cntr2[0]))
Previous code.
It seems like this code does what you want. It treats the case of real inner tangents of disjoint ellipses or the case of two complex tangents of two ellipses, intersecting at two real points. For circles, the result seems to match the elementary formula.
'''
angle between inner tangents of two non-intersecting ellipses
or the complex angle between the two complex tangents of two
intersecting ellipses at two real points
'''
import numpy as np
import math
import cmath
'''
####### FUNCTIONS IMPORTANT FOR THE CALCULATION #######
'''
def homogenize(x):
return np.array([x[0], x[1], 1])
# transforming the vector of six coefficients
# of a quadratic equation with two variables into a matrix of
# the corresponding conic
def conic_from_equation(eq):
'''
eq is an np.array vector of the equadratic equation's six coefficients
eq[0]*x**2 + eq[1]*x*y + eq[2]*y**2 + eq[3]*x + eq[4]*y + eq[5] = 0
conc is a symmetric matrix of the bilinier form of the quadratic equation
'''
return np.array([[2*eq[0], eq[1], eq[3]],
[ eq[1], 2*eq[2], eq[4]],
[ eq[3], eq[4], 2*eq[5]]]) / 2
# transforming the matrix of a conic into a vector of six coefficients
# of a quadratic equation with two variables
def equation_from_conic(C):
'''
c[0]*x**2 + c[1]*x*y + c[2]*y**2 + c[3]*x + c[4]*y + c[5] = 0
'''
return np.array([C[0,0], 2*C[0,1], C[1,1], 2*C[0,2], 2*C[1,2], C[2,2]])
# obtaining the conic dual to a given conic,
# both represented as symmetric 3x3 matrices
def dual_conic(conic):
return np.linalg.inv(conic)
# given a pair of conics, as a pair of symmetric 3x3 matrices,
# calculate the vector k = (k[0], k[1], k[2]) of values for each of which
# the conic c1 - k[i]*c2 from the pencil of conics c1 - t*c2
# is a degenerate conic (the symmetric tensor product of a pair of linear forms)
# and also find the matrix U
# of the projective transformation that simplifies the geometry of
# the pair of conics, the geometry of the pencil c1 - t*c2 in general,
# as well as the geometry of the three degenerate conics in particular
def trivialization_of(conic1, conic2):
'''
c1 and c2 are 3 by 3 symmetric matrices of two conics (possibly dual)
'''
c21 = np.linalg.inv(conic2).dot(conic1)
k, U = np.linalg.eig(c21)
return k, U
# find the common points, i.e. points of intersection of a pair of conics
# represented by a pair of symmetric matrices
def find_intersection_points(c1, c2):
k, U = trivialization_of(c1, c2)
L1 = (U.T).dot((c1 - k[0]*c2).dot(U))
L2 = (U.T).dot((c1 - k[1]*c2).dot(U))
acc = 1e-12
are_complex = (k.imag > acc).any()
if are_complex:
x_0 = cmath.sqrt(-L2[2,2] / L2[0,0])
y_0 = cmath.sqrt(-L1[2,2] / L1[1,1])
else:
x_0 = math.sqrt(abs(L2[2,2] / L2[0,0]))
y_0 = math.sqrt(abs(L1[2,2] / L1[1,1]))
sol = np.array([[ x_0, y_0, 1],
[-x_0, y_0, 1],
[-x_0, -y_0, 1],
[ x_0, -y_0, 1]])
sol = sol.dot(U.T)
return sol, are_complex
# find the three coefficients of all four common tangents to a pair of conics
# make sure that the conics do have four common tangents,
# which is the generic situation
def find_common_tangents(c1, c2):
dc1 = dual_conic(c1)
dc2 = dual_conic(c2)
return find_intersection_points(dc1, dc2)
# if the two conics are ellipses, they do not intersect the line at infinity.
# by polarity, the poles of the line at infinity are
# the (affine and in fact metric) centers of the ellipses and are located
# inside the ellipses each. We use them to detect the inner tangents and
# to orient the vectors normal to these tangents in a way
# so that the angle between these two normal vectors is the vector on the picture
def dual_to_infinity_line(conic):
p = dual_conic(conic).dot(np.array([0,0,1]))
return homogenize( p[0:2] / p[2] )
# find the common inner tangents between a pair of conics,
# represented as 3x3 symmetric matrices, with properly oriented normal vectors
def find_common_inner_tangents(c1, c2):
tn, complex_tangents = find_common_tangents(c1, c2)
if not complex_tangents:
p1 = dual_to_infinity_line(c1)
p2 = dual_to_infinity_line(c2)
# boolean vector, true if inner tangent false if not
is_inner_tangent = ( tn.dot(p1)*tn.dot(p2) < 0 )
tn = tn[ is_inner_tangent, : ]
# reorient one of the normal vectors so that they give us the proper angle
if tn[0,:].dot(p1)*tn[1,:].dot(p1) > 0:
tn[1,:] = - tn[1,:]
else:
acc = 1e-12
is_complex_line = np.logical_or(abs(tn[:,0].imag) > acc,
abs(tn[:,1].imag) > acc,
abs(tn[:,2].imag) > acc)
tn = tn[ is_complex_line, :]
return tn, complex_tangents
# calculate the angle between a pair of line, given by their three coefficients
def angle_bn_R_lines(line1, line2):
theta = line1[0:2].dot(line2[0:2]) / math.sqrt( line1[0:2].dot(line1[0:2])*line2[0:2].dot(line2[0:2]))
return math.acos(theta) #*180/np.pi
def angle_bn_C_lines(line1, line2):
theta = line1[0:2].dot(line2[0:2]) / ( cmath.sqrt( line1[0:2].dot(line1[0:2]))*cmath.sqrt(line2[0:2].dot(line2[0:2])) )
return cmath.acos(theta)
# calculate the proper angle between the inner or complex tangents of two quadratic equations
# that represent a pair of conics, hopefully non-intersecting
def angle_bn_inner_tangents_of(eq_of_conic1, eq_of_conic2):
c1 = conic_from_equation(eq_of_conic1)
c2 = conic_from_equation(eq_of_conic2)
tn, complex_tangents = find_common_inner_tangents(c1, c2)
if complex_tangents:
return angle_bn_C_lines(tn[0,:], tn[1,:])
else:
return angle_bn_R_lines(tn[0,:], tn[1,:])
'''
####### END OF FUCTIONS IMPORTANT FOR THE CALCULATION #######
'''
'''
####### auxiliary functions to handle the test scenario:
'''
def cos_sin(angle_deg):
return math.cos(angle_deg*math.pi/180), math.sin(angle_deg*math.pi/180)
def rotation(cs_sn):
return np.array([[cs_sn[0], -cs_sn[1]],
[cs_sn[1], cs_sn[0]]])
def isom(angle, translation):
'''
isometry from global coordinate system (world system)
to conic-aligned coordinate system (conic attached)
'''
cos_, sin_ = cos_sin(-angle)
tr = - rotation((cos_, sin_)).dot(translation)
return np.array([[ cos_, -sin_, tr[0]],
[ sin_, cos_, tr[1]],
[ 0, 0, 1 ]])
# calculating the conic defined by a pair of axes' lengths,
# axes rotation angle and center of the conic
def Conic(major, minor, angle, center):
D = np.array([[minor**2, 0, 0],
[ 0, major**2, 0],
[ 0, 0, -(minor*major)**2]])
U = isom(angle, center)
return (U.T).dot(D.dot(U))
'''
####### end of auxiliary functions
'''
#%%
'''
Example of finding the angle between the common tangents of a pair of conics:
conic 1 (circle): major axis = 2, minor axis = 2, angle = 0, center = (0,0)
conic 2 (circle): major axis = 3, minor axis = 3, angle = 0, center = (4,0)
'''
a = 2
b = 2
cntr = np.array([0,0])
w = 0
Eq1 = equation_from_conic(Conic(a, b, w, cntr))
Co1 = conic_from_equation(Eq1)
a = 3
b = 3
cntr = np.array([4,0])
w = 0
Eq2 = equation_from_conic(Conic(a, b, w, cntr))
Co2 = conic_from_equation(Eq2)
four_tngs = find_common_tangents(Co1, Co2)
#inner_tngs = find_common_inner_tangents(Co1, Co2)
print( angle_bn_inner_tangents_of(Eq1, Eq2) )
print(2*cmath.asin((2+3)/4))
$$$$
Older Post.
Today I had some time to experiment with the code I wrote in the previous post. Theoretically, the approach should be exactly the same as the case of four real tangents. However, the only major hassle I am seeing for now is the complex arithmetic. In the function find_common_points() some square roots are calculated, which works fine for real numbers, but when the entries of the trivialized degenerate conics L1 and L2 become complex (and they do in the case of two ellipses intersecting at exactly two points), then one needs to carefully calculate the appropriate complex square root. The rest is the same. With two complex tangents and two real ones, you choose the complex to be "inner" and calculate a complex dot product divided by complex norms, which will give you a complex number. After that you can take complex arccos() of that. Whether this scheme leads to the same elementary circle calculation you show in your post, is unclear for now, although the fact that all of these are complex holomorphic functions of several variables that overlap in the real case suggest that it is possible these constructions match due to the rigidity of holomorphic prolongations.
Best Answer
Using pythagoras theorem and similarity of triangles, you can easily find the other radius which is $4$ and distance between centres is $5$.
Now let centre of circle with radius $r_1=3$ be $C_1$ and with radius $r_2=4$ be $C_2$. Let intersection of direct tangent and line through $C_1C_2$ be $P$.
Using similarity,
$$\frac{PC_1}{3} = \frac{PC_1+C_1C_2}{4}= \frac{PC_1+5}{4}$$ This gives $PC_1 = 15$. So now let half of the required angle be $\alpha$, this is the angle between direct tangent and line through $C_1C_2$.
$$\sin \alpha = \frac{r_1}{PC_1} = \frac{1}{5},\,\, \cos\alpha = \frac{2\sqrt{6}}{5}$$
So required answer is $\sin 2\alpha = \frac{4\sqrt{6}}{25}$