A locus is the set of all points that satisfy a given condition or a set of given conditions.
All examples here deal about points in a plane.
In the examples below we work with an orthonormal coordinate system.
The choice of this coordinate system is very important because the difficulty of the calculations
depends strongly on this choice.
Find the locus of the points equally distant from two intersecting lines a and b.
We choose the coordinate system so that a and b are symmetric relative to the x-axis.
The lines a and b have equations y = m x and y = - m x.
P(x,y) is a point on the locus
<=>
|P,a| = |P,b|
<=>
|P,a|2 = |P,b|2
<=>
(y - m x)2 (y + m x)2
------------ = --------------
1 + m2 1 + m2
<=>
(y - m x)2 = (y + m x)2
<=>
x.y = 0
<=>
x= 0 or y = 0
Consider the hyperbola y = 1/x. Through a variable point P (x, y) of the plane, we draw a line parallel to the x-axis
and another line parallel to the y-axis. These lines intersect the hyperbola respectively in A and B.
Find the locus of point P such that the centroid of triangle PAB is on the line x= 1.
The coordinates of P are (x,y)
The coordinates of A are (x, 1/x)
The coordinates of B are (1/y, y)
The x-value of the centroid is ( x + x + 1/y )/3 = (2x+ 1/y)/3
P(x,y) is a point on the locus if and only if (2x+ 1/y)/3 = 1
The locus has equation y = 1/(-2x+3).
It is the equation of a homographic function.
It is a hyperbola with asymptotes y=0 and x=3/2.
You can find more info abaut the homographic function via this link .
A triangle ABC has a fixed side [AB] with length c.
Find the locus of the third vertex C such that
the medians from A en C are orthogonal.
Next, find the maximum area of such a triangle.
Analytical
We choose an orthonormal coordinate system such that A(-c/2,0), B(c/2,0).
C(x,y) is the variable third vertex.
Make a figure.
The center of [BC] is M( (2x+c)/4, y/2 ).
The median from C has a slope y/x.
The median AM has a slope 2y/(2x+3c).
C(x,y) is a point of the locus
<=>
The medians from A and C are orthogonal
<=>
y 2y
---. ----------- = -1
x 2x + 3c
<=>
2 y2 + 2x2 + 3c x = 0
<=>
x2 + y2 + (3c/2) x = 0
<=>
(x + 3c/4)2 + y2 = 9c2/16
The locus of the vertex C is a circle
with center (-3c/4,0) and radius 3c/4
The area is maximum if and only if the height of the triangle is maximum.
The maximum height is 3c/4. The maximum area is 3c2/8.
Geometric:
Say Z is the centroid of the variable triangle ABC.
A and B are fixed points.
The distance|AB| = c. Point O is the center of [AB].
C(x,y) is a point of the locus
<=>
The medians AZ and CZ are orthogonal
<=>
The triangle AOZ is a right-angled triangle
The fixed segment [AO] is the hypotenuse of the variable right-angled triangle AOZ.
So, the locus of Z is the circle C1 with center (-c/4,0). The radius is c/2.
But Z is the centroid of the triangle. So, the vector C = 3Z.
From this, it follows that the locus of point C is a circle C2. It is the image
of the transformation of circle C1 by a homothety with center O and factor 3.
The center of the circle C2 is (-3c/4,0) and the radius is 3c/2.
m is a variable parameter.
For each fixed value of m, the plots P and b are fixed.
Therefore, we say that P and b are associated curves.
If m changes smoothly, the curves P and b change smoothly.
The intersection points of P and b will also change smoothly.
The intersection points describe a curve c. c is the locus.
P(x,y) is a point of c
<=> P(x,y) is an intersection point of P and b
<=> There is a value of m such that
/ y = x2 + mx
\ y = 2x + m
<=> The following system has a solution for m
/ mx = y - x2
\ m = y - 2x
(elimination theory)
<=>
| 1 y-2x |
| | = 0
| x y - x2 |
<=> x2 - xy + y = 0
If we omit the intermediate calculations here, we see better what we found:
P(x,y) is a point of c
<=> x2 - xy + y = 0
The last equation is the equation of the locus c.
The equation of the locus arises by eliminating the parameter from the system
formed by the associated curves.
The real numbers l and m are parameters, not both 0.
If we replace l and m with r.l and r.m the lines does not change.
Because of this property, we say that l and m are homogeneous parameters.
The lines a and b are associated lines.
The intersection point describes the locus c.
P(x,y) is on c
<=> P(x,y) is an intersection point of a and b
<=> There is a value of l and m ( not both 0) such that
l x - m y = 0
m x + l y = 5l
<=> The following system has a solution for l and m ( not both 0)
x l - y m = 0
(y - 5) l + x m = 0
(elimination theory)
<=> | x -y |
| | = 0
|y - 5 x |
<=> x2 + y2 - 5 y = 0
The last equation is the equation of the locus c. The locus is a circle. The associated lines intersect on this circle.
Each point of the circle is an intersection point of the two variable lines.
The equation of the locus arises by eliminating the parameters from the system
formed by the associated lines.
Take a circle with radius 2. Consider a variable chord with length 2 of this circle.
Find the locus of the center of this chord.
By geometric thinking you can easily determine the locus. We will calculate the locus here
in order to work with trigonometric functions and to illustrate a special elimination.
Take center of the circle in O (0.0).
Since the chord [AB] is as large as the radius, the included angle between OA and OB is pi/3 radians.
A(2.cos(t), 2.sin(t)) is a variable point of the circle.
Then, we have B(2.cos(t+pi/3), 2.sin(t+pi/3) )
The center M of the chord has coordinates
x = ( cos(t) + cos(t+pi/3))
y = ( sin(t) + sin(t+pi/3))
With Simpson's formulas ...
x = 2.cos(t+pi/6) . cos(pi/6)
y = 2.sin(t+pi/6) . cos(pi/6)
we simplify this to
x = sqrt(3).cos(t+pi/6)
y = sqrt(3).sin(t+pi/6)
These are actually the equations of two lines. These lines are the associated lines.
They intersect in the middle of the variable chord.
P(x,y) is a point of the locus
<=>
There is a t-value such that
x = sqrt(3).cos(t+pi/6)
y = sqrt(3).sin(t+pi/6)
<=>
the following system has a solution for t
cos(t+pi/6) = x/ sqrt(3)
sin(t+pi/6) = y/ sqrt(3)
(elimination theory)
<=>
x2/3 + y2/3 = 1
<=>
x2 + y2 = 3
The locus is a circle, as expected.
The equation of the locus arises by eliminating the parameter from the system
formed by the associated lines.
The points A (a, 0) and B (b, 0) are fixed points.
A point C is variable along the y-axis.
Point D is the mirror image of C relative to the x-axis.
Find the locus of the intersection of AC en BD.
AC and BD are the associated lines
Take C(0,m), then D(0,-m). m is variable.
The equation of AC is mx + ay - am = 0.
The equation of BD is -mx + by + bm = 0.
P(x,y) is intersection point of AC and BD
<=>
There is an m such that
/ mx + ay - am = 0
\ -mx + by + bm = 0
<=>
The following system has a solution for m
/ (x - a)m = -ay
\ (b - x)m = by
(elimination theory)
<=>
| x - a -ay|
| b - x by| = 0
<=>
y = 0 or x = 2 ab /(a + b)
The part y=0 arises at the moment that D = C.
The most interesting part of the locus is the line x = 2 ab /(a + b).
The equation of the locus arises by eliminating the parameter from the system
formed by the associated lines.
Consider the hyperbola with equation y=1/x. Through a variable point P of the plane one takes a line parallel
to the x-axis and a line parallel to the y-axis. These lines intersect with the hyperbola in the points A
an B respectively. Find the locus of point P such that the centroid of the triangle PAB is on the line x=1.
Construct a figure with all the items on it. We choose variable coordinates (r,s) for P. Point P,
which describes the locus, is the intersection point of the associated lines x=r and y=s. At first
sight, there are two parameters r and s. But r and s are connected through a condition.
Point A is A(1/s,s) and point B is B(r,1/r). The centroid of the triangle PAB is Z((2r+1/s)/3 , (2s+1/r)/3).
Z is on the line x=1 if and only if (2r+1/s)/3 = 1. This is equivalent to 2rs-3s+1=0.
This is the condition that gives the connection between the parameters r and s. This equation 2rs-3s+1=0
is called a binding equation.
The equation of the locus arises by eliminating the parameter r an s from the system
formed by the associated lines and the binding equation.
x = r
y = s
2 r s - 3 s + 1 = 0.
We consider r and s as unknowns and we build the condition such that the system has a solution for r and s.
r = x
s = y
2 r s - 3 s = -1
The condition is 2 x y - 3 y + 1 = 0.
This is the equation of the locus. It can also be written in the form y = 1/(3-2x).
The graph of the locus is a hyperbola.
We start with a circle C with equation x2 + y2 = 1 and two fixed tangent lines p and q
with equation x=1 and y=1.
A variable tangent line r to the circle intersects p in P and q in Q. Find the locus
of the intersection point S of the line a through P parallel to the x-axis and the line b
through Q parallel to the y-axis.
We take a variable point V(cos(t),sin(t)) on the circle. The tangent line in V to the circle is
x.cos(t)+y.sin(t)=1.
The coordinates of P are (1, (1-cos(t))/sin(t))
The coordinates of Q are ( (1-sin(t))/cos(t) , 1)
The line a is y = (1-cos(t))/sin(t)
The line b is x = (1-sin(t))/cos(t)
The point S of the locus is the intersection point of the two associated lines a and b.
The equation of the locus arises by eliminating the parameter t between the two equations
of the associated lines.
We solve the system [x = (1-sin(t))/cos(t) ; y = (1-cos(t))/sin(t)] for sin(t) and cos(t).
sin(t) = (x - 1)/(x y - 1)
cos(t) = (y - 1)/(x y - 1)
Now we require that cos2(t) + sin2(t) = 1
After substitution, we bring all the non-zero terms to the left side and after factoring we obtain:
(x - 1) (y -1) (x y + x + y - 1)=0
The locus can be decomposed into three parts
First we consider the curve x y + x + y - 1 =0
This is a hyperbola with equation y = (1-x)/(1+x)
The horizontal asymptote is y = -1 and the vertical asymptote is x = -1
The second part is the line x = 1.
If point V coincides with the point (1,0) then point P is an arbitrary point on the line
x = 1. Then point Q is point(1,1). The intersection point S is an arbitrary point on x = 1.
The line x = 1 is actually not a true set of intersection points of a and b.
That's why this line is called a parasitic part of the locus.
The third part is the line y = 1.
If point V coincides with the point (1,1) then PQ is the line y = 1 and P is (1,1).
Also this line is not a real collection of intersection points of a and b.
It is also a parasitic part of the locus.
In an orthonormal coordinate system, a circle has equation x2+y2=1
and A(2,0) is a fixed point.
A variable line through point A, intersects the circle in points B and C.
Calculate the locus of midpoint M of the segment [BC].
The variable line is y = m (x - 2).
The points B and C are the
solutions of the system
x2 + y2 = 1
y = m (x - 2)
The x-values are the roots x1 and x2 of
(1 + m2) x2 - 4 m2 x + 4 m2 - 1 = 0
The y-values are the roots y1 and y2 of
(1 + m2) y2 + 4 m y + 3 m2 = 0
The x-value of M is (x1 + x2)/2 = 2 m2/(1 + m2)
The y-value of M is (y1 + y2)/2 = - 2 m /(1 + m2)
So, we can say that M is the intersection point of the
associated lines
x = 2 m2/(1 + m2) (1)
y = - 2 m /(1 + m2) (2)
The locus arises from eliminating m from these equations.
(1) <=> (x - 2) m2 + x = 0 (3)
(2) <=> y m2 + 2 m + y = 0 (4)
We eliminate m.
We ensure that m2 occurs only in one of the equations.
We calculate y.(3) - (x-2).(4) and after simplification, we have
(x - 2) m2 + x = 0 (3)
(x-2) m - y = 0 (5)
We now bring m from the second equation in the first equation.
After simplification, we have
x2 + y2 - 2x = 0
<=>
(x - 1)2 + y2 = 1
The locus is a circle with midpoint (1,0) and radius = 1.
But, when you draw several instances of the variable line through
point A, there are many positions of that line that don't generate
points B and C. From this we see that the locus is only that part
of the circle that is inside or on the circle x2+y2=1.
Therefore, we say that all the other points of the locus are
parasitic points.
C is a circle with center A(1,0) and radius one and C' is a circle with center
A'(-1,0) and radius one.
Through the origin O(0,0) we draw a variable line that intersects C' in a second point B'.
Find the locus of the intersection point P of the line A'B' and the line through A orthogonal to
the line OB'.
You can find the locus by giving the equation y=mx to the line OB'.
Then you have to eliminate the parameter m between the equations of the lines AP and A'B'.
We will look into this problem with different approach. We start with a figure.
Let Q be the intersection point of B'P and the circle C'.
OQ is orthogonal to OB' because B'Q is a diameter of the circle C'.
Construct A'R perpendicular to OB'.
Now we have A'R // QO // PA (1)
Let S be the intersection point of AP and OB'.
The triangle ORA' is congruent with the triangle OSA.
So, |O,R| = |O,S| (2)
From (1) and (2)
1 = |A',Q| = |Q,P|
So, |A',P| = 2 = a fixed distance
The locus is a circle with center A' and radius 2.
Sometimes geometry is stronger than analysis. But the cooperation of both is brilliant!
The numbers a and c are constants and a > c > 0.
We start with two points F(c,0) and F'(-c,0) and we take a circle C with center F and
a variable radius m. Now we take a second circle C' with center F' and a variable radius
(2a - m).
The sum of the radii of C and C' is the constant number 2a.
We consider the two circles as associated curves and we'll calculate the
locus of the intersection of the two curves.
C has equation (x - c)2 + y2 = m2
C' has equation (x + c)2 + y2 = (2a - m)2
We find the locus when we eliminate the parameter m from the system
But that parameter m appears quadratically in each equation of the system.
Using the first equation we ensure that m2 disappears from the second equation.
The line d is a variable line through the fixed point A (a, 0)
and that line d cuts the y-axis in point S. The perpendicular from S on d
intersects the x-axis in point B.
Find the locus of the mirror image P of point B relative to the line d.
The line d has equation y = m(x-a). Here m is the parameter. Now we have
S(0, -a m)
Line BS has as equation y + a m = (-1/m) x
Then we find B(-am2, 0) and P(a m2, -2 a m)
Now, we know the location of a variable point P of the locus,
but we don't know the equation of the locus.
At first glance, we see no associated lines.
But based on the coordinates of P, we can write:
The lines x = a m2 and y = -2 a m are intersecting in P and
we can use these lines as associated lines.
Now it is sufficient to eliminate m from the system
x = a m2
y = -2 a m
One find a y2 = 4 x.
That curve is a parabola with the x-axis as symmetry axis.
The small circle start rolling in point (R,0). Then t=s=0.
Since the circle is rolling without slipping, we have
that the two blue arcs have the same length.
So, R.t = -R.s/4 <=> s = - 4 t
P = A + AP
=>
P.E1 = A.E1 + AP.E1
<=>
x = (3R/4).1.cos(t) + (R/4).1.cos(t+s)
<=>
x = (3R/4).cos(t) + (R/4).cos(3t) (1)
and in the same way
P.E2 = A.E2 + AP.E2
=>
y = (3R/4).1.cos(pi/2 -t) + (R/4).1.cos(t - pi/2 + s)
<=>
y = (3R/4).sin(t) + (R/4).sin(3t) (2)
Since cos(3t) = 4 cos3(t) - 3cos(t)
sin(3t) = 3 sin(t) - 4 sin3(t)
we find the parametric equations of the astroid from (1) and (2)
x = R cos3(t)
y = R sin3(t)
The cartesian equation can be found through elimination of t.
The parametric equations of an astroid are x= cos3(t) y=sin3(t).
Find the locus of the point P such that the tangent lines from P
to the astroid are orthogonal.
First we take a variable point A(cos3(t), sin3(t)) on the astroid.
The slope of the tangent line in A is
dy 3 sin2(t) cos(t)
---- = ------------------ = - tan(t)
dx -3 cos2(t) sin(t)
Now, we take a variable point B(cos3(t'), sin3(t')) on the astroid.
The slope of the tangent line in B is - tan(t')
The two tangent lines are orthogonal if and only if
tan(t'). tan(t) = -1
<=> tan(t') = - 1/tan(t)
<=> tan(t') = tan(t+ pi/2)
<=> t' = t + pi/2 + k pi
It is sufficient to study the cases
t' = t + pi/2 en t' = t + 3pi/2
since the other values of t bring anything new.
The equation of the tangent line in A is
y - sin3(t) = -tan(t) ( x - cos3(t))
<=> y = -tan(t).x + sin(t) cos2(t) + sin3(t)
<=> y = -tan(t).x + sin(t) (1)
The equation of the tangent line in the point with t'= t + pi/2 is
y = -tan(t').x + sin(t')
<=> y = cot(t).x + cos(t) (2)
The equation of the tangent line in the point with t'= t + 3pi/2 is
y = -tan(t').x + sin(t')
<=> y = cot(t).x - cos(t) (3)
The tangent lines (2) and (3) are both orthogonal to the tangent (1).
First Case:
We start with the tangent lines (1) and (2)
y = -tan(t).x + sin(t) (1)
y = cot(t).x + cos(t) (2)
We calculate the point of intersection of these two tangents.
We solve this system for x and y and we find after simplification
x = (1/2) sin(2t) ( sin(t) - cos(t) ) (4)
y = (1/2) sin(2t) ( sin(t) + cos(t) ) (5)
This gives us for each t the point from which the tangents
to the asteroid are orthogonal lines.
(4) an (5) are the parametric equations of the locus.
The the locus is indicated in blue.
Second Case:
We take the tangent (1) and (3). Now, one finds than for x and y
x = (1/2) sin(2t) ( sin(t) + cos(t) )
y = (1/2) sin(2t) ( cos(t) - sin(t) )
We'll show that this is the same locus as in the first case.
Since t is a real parameter, the shape of the locus does not change if we replace t by
t + pi/2.
Now, it is clear that this locus is the same one as in the first case.
Note:
The locus is tangent to the asteroid in eight points. Let C be such a point.
The two tangent lines from C to the asteroid are orthogonal lines.
Since C is on the asteroid the first tangent is the tangent in C to the astroid.
The other tangent is a normal line in C.
Conclusion:
The eight special points are the only points of the astroid such that the normal line
in these points are also tangent lines to the astroid.
The given solution is not 'the' solution.
Most exercises can be solved in different ways.
It is strongly recommended that you at least try to solve the
problem before you read the given solution.
Find all points P(x,y) such that sin2(x) = cos2(y)
sin2(x) = cos2(y)
<=>
sin(x) = cos(y) or sin(x) = - cos(y)
<=>
cos(pi/2 -x) = cos(y) or cos(pi/2 -x) = cos(pi - y)
<=>
pi/2 -x = ± y + 2k pi or pi/2 -x = ±(pi - y) + 2k pi
<=>
y = ± (x - pi/2) + 2k pi or y = ±(x + pi/2) + 2k pi
<=>
y = ± x + pi/2 + k pi
The locus consists of a network of lines.
Point A(0,1) is fixed and point P(p,0) is variable.
The perpendicular to AP through point P intersects the y-axis in point M.
Find the locus of the point Q such that P is the center of the segment MQ.
Sketch a figure.
In the right triangle APM is the altitude |OP| to the hypotenuse the mean proportional between the segments |OA| and |OM|, into which it divides the hypotenuse.
|OP|2 = |OA|.|OM| => p2 = 1.|OM|
so, M = M(0, -p2 ) and Q = Q(2p, p2).
The variable lines x = 2p and y=p2 intersect in Q. They are associated lines.
p is the parameter.
We find the equation of the locus by eliminating the parameter p.
It is the equation of the parabola y = x2/4.
A circle C1 with radius 3 has its center in O(0,0) and a circle C2
with radius 1 has its center in D(6,0).
Draw a tangent from a point P to C1 and C2. The tangent points are A and B
respectively.
Find the locus of point P such that |PA|2 +|PB|2 = 58.
P(x,y) is a point of the locus
<=>
|PA|2 +|PB|2 = 58
<=>
|PO|2 -9 + |PD|2 -1 = 58
<=>
x2 + y2 + (x-6)2 + y2 -10 = 58
<=>
...
<=>
(x-3)2 + y2 = 25
The locus is a circle with center (3,0) and radius 5.
The points of this circle that are inside the given circles are parasitic points
of the locus.
A circle C1 with radius R=5 has its center in O(0,0) and a circle C2
with radius r = 3 has its center in A(10,0).
Draw a tangent from a point P to C1 and C2. The tangent points are D and E
respectively.
Find the locus of point P such that |PD| = |PE|.
Hint: show first that P is a point of the locus if and only if
|PO|2 - |PA|2 = R2 -r2.
Make a figure.
|PO|2 = |PD|2 + R2
|PA|2 = |PE|2 + r2
P(x,y) is a point of the locus
<=>
|PD|2 = |PE|2
<=>
|PO|2 - R2 = |PA|2 - r2
<=>
|PO|2 - |PA|2 = R2 - r2
<=>
x2 + y2 - (x - 10)2 - y2 = 16
<=>
...
<=>
20 x = 116
<=>
x = 5.8
The locus is a line perpendicular to OA.
A circle C1 with radius R = 5 has its center in A(-5,0) and a circle C2
with radius r = 3 has its center in B(3,0).
A variable line d1 through O intersects C1 a second time in point D.
Through O we draw the line d2 perpendicular to d1. The line d2
intersects C2 a second time in point E.
Find the locus of the center M of [DE].
Hint: Sometimes geometry is stronger than calculations.
First we recall two geometric properties.
If a line is drawn parallel to one side of a triangle so that it intersects the other two sides, it divides them proportionally.
The perpendicular bisector of a chord of a circle passes through the center.
The perpendicular bisector of [OD] contains A and M.
The perpendicular bisector of [OE] contains B and M.
The triangle AMB is a variable right angled triangle with hypotenuse [AB].
The locus of point M is a circle with diameter [AB].
It is also possible to find this locus by calculations.
On a circle with radius r and center O(0,0) we take the points B an C with
ordinate (-r/2).
Point P is a variable point on the circle. Find the locus of the
centroid of the triangle PBC.
The vertices of the triangle are
P( r cos(t), r sin(t) )
B( - k , -r/2 )
C( k , -r/2 )
The coordinates of the centroid are
r cos(t) r sin(t) - r
( --------- , -------------- )
3 3
This centroid is always the intersection of the associated lines
r cos(t)
x = -----------
3
r sin(t) - r
y = --------------
3
The equation of the locus results from the elimination of the parameter t.
/ 3 x = r cos(t)
\ 3y +r = r sin(t)
(3x)2 + (3 y+ r)2 = r2
<=>
x2 + (y + r/3)2 = (r/3)2
The locus is a circle with center M(0, -r/3) and radius r/3.
On a circle with radius r and center O(0,0) we take the points
A = A(r,0) and B = B(0,r).
Point C is a variable point on the circle.
Find the locus of point D such that (A,D) and (B,C) are
the pairs of opposite vertices of a parallelogram.
Let D = D(x,y). C = C(a,b) is a variable point on the circle.
The parameters a and b are
connected with the binding equation a2 + b2 = r2.
Let M be the intersection point of the diagonals of the parallelogram.
Then M is the midpoint of [BC] and so M = M(a/2 , (b+r)/2 ).
And M also the midpoint of [AD], M = M( (x+r)/2 , y/2 ).
From this we have
/ a = x+r
\ b + r = y
<=>
/ a = x+r
\ b = y-r
The parameters a and b are eliminated using the binding equation.
(x + r)2 + (y - r)2 = r2
The locus is a circle with center (-r, r) and radius r.
The point A(0,r) is a fixed point on the circle with center O(0,0).
The point P is a variable point on the circle.
Point M is the center of [OP] and point N is the center of [AM].
Find the locus of point N.
Draw a clear figure.
P(r cos(t), r sin(t))
Then
M((r/2) cos(t) , (r/2) sin(t))
and
N( (r/4)cos(t) , (1/2)( r + (r/2)sin(t))
<=>
N( (r/4)cos(t) , (r/2)+(r/4)sin(t) )
N is the intersection of the associated lines
/ x = (r/4)cos(t)
\ y = (r/2)+(r/4)sin(t)
<=>
/ x = (r/4)cos(t)
\ y - r/2 = (r/4)sin(t)
We eliminate t by squaring those two equations and add them.
x2 + (y - r/2)2 = (r/4)2
The locus is the circle with center (0,r/2) and radius r/4.
On the circle with equation x2 + (y-1)2 = 2 , we take two fixed points
B(1,0) and C(-1,0) and a variable point A.
Find the locus of the orthocenter of the triangle ABC.
Draw a clear figure.
We give A the coordinates (xo, yo).
Since A is on the circle, xo and yo are connected by the binding equation
As associated lines we take the altitudes through A and B.
The altitude through A is
x = xo
The altitude through B is
xo + 1
y = - -------- (x-1)
yo
To obtain the equation of the locus, we have to eliminate the
parameters xo and yo from the system
/
| x = xo
| xo + 1
| y = - -------- (x-1)
\ yo
We simplify the second equation, based on the first one.
<=>
/
| x = xo
| xo + 1
| y = - -------- (xo-1)
\ yo
<=>
/
| x = xo
| xo2 - 1
| y = - -----------
\ yo
<=>
/
| x = xo
| 1 - xo2
| y = -----------
\ yo
Relying on the binding equation (**), we can further simplify
/
| x = xo
| yo2 - 2 yo
| y = ---------------
\ yo
<=>
/
| x = xo
|
| y = yo - 2
\
<=>
/
| xo = x
|
| yo -1 = y + 1
\
Now, we eliminate xo and yo
Relying on the binding equation (*) , we find
x2 + (y +1)2 = 2
This circle is the locus of the orthocenter.
It is the reflection of the given circle relative to the line BC.
Send all suggestions, remarks and reports on errors to Johan.Claeys@ping.be
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Let Q be the intersection point of B'P and the circle C'.
OQ is orthogonal to OB' because B'Q is a diameter of the circle C'.
Construct A'R perpendicular to OB'.
Now we have A'R // QO // PA (1)
Let S be the intersection point of AP and OB'.
The triangle ORA' is congruent with the triangle OSA.
So, |O,R| = |O,S| (2)
From (1) and (2)
1 = |A',Q| = |Q,P|
So, |A',P| = 2 = a fixed distance
