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.
Take two fixed points A and B. Find the locus of points P such that PA = 3.PB.

P(x,y) is a point on the locus <=> PA = 3.PB <=> PA^{2} = 9.PB^{2} <=> (x1)^{2} + y^{2} = 9 ((x + 1)^{2} +y^{2}) <=> ... <=> 2 x^{2} + 2 y^{2} + 5 x + 2 = 0The last equation is an equation of a circle. Each point of the circle has the requested property, and vice versa.
Find the locus of the points equally distant from two intersecting lines a and b.

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 + m^{2} 1 + m^{2} <=> (y  m x)^{2} = (y + m x)^{2} <=> x.y = 0 <=> x= 0 or y = 0We find, as expected, xaxis and yaxis as locus
Point A is a fixed point A(a,0). P is a variable point in the plane. P 'is the mirror image of P relative to the yaxis. (draw a figure) Find the locus of point P such that PA^{2} + P'A^{2} = 4a^{2}.

P(x,y) is a point on the locus <=> PA^{2} + P'A^{2} = 4 a^{2} <=> (x  a)^{2} + y^{2} + (x  a)^{2} + y^{2} = 4 a^{2} <=> ... <=> x^{2} + y^{2} = a^{2}
Consider the hyperbola y = 1/x. Through a variable point P (x, y) of the plane, we draw a line parallel to the xaxis and another line parallel to the yaxis. 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 xvalue 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. 
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 y^{2} + 2x^{2} + 3c x = 0 <=> x^{2} + y^{2} + (3c/2) x = 0 <=> (x + 3c/4)^{2} + y^{2} = 9c^{2}/16 The locus of the vertex C is a circle with center (3c/4,0) and radius 3c/4The area is maximum if and only if the height of the triangle is maximum. The maximum height is 3c/4. The maximum area is 3c^{2}/8.
Say Z is the centroid of the variable triangle ABC.
A and B are fixed points.
The distanceAB = 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 rightangled triangleThe fixed segment [AO] is the hypotenuse of the variable rightangled triangle AOZ.
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 C_{2}. It is the image
of the transformation of circle C_{1} by a homothety with center O and factor 3.
The center of the circle C_{2} is (3c/4,0) and the radius is 3c/2.
It is assumed that you are aware of the elimination methods. These methods are explained here.
/ y = x^{2} + mx \ y = 2x + mm is a variable parameter.
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 = x^{2} + mx \ y = 2x + m <=> The following system has a solution for m / mx = y  x^{2} \ m = y  2x (elimination theory) <=>  1 y2x    = 0  x y  x^{2}  <=> x^{2}  xy + y = 0If we omit the intermediate calculations here, we see better what we found:
P(x,y) is a point of c <=> x^{2}  xy + y = 0The 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.
/ l x  m y = 0 \ m x + l y = 5lThe 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.
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  <=> x^{2} + y^{2}  5 y = 0The 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.

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 tvalue 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) <=> x^{2}/3 + y^{2}/3 = 1 <=> x^{2} + y^{2} = 3The 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. 
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 xaxis and a line parallel to the yaxis. 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. 
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 2rs3s+1=0. This is the condition that gives the connection between the parameters r and s. This equation 2rs3s+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 = 1The 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/(32x). The graph of the locus is a hyperbola.
We start with a circle C with equation x^{2} + y^{2} = 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 xaxis and the line b through Q parallel to the yaxis. 
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.
We solve the system [x = (1sin(t))/cos(t) ; y = (1cos(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 cos^{2}(t) + sin^{2}(t) = 1
After substitution, we bring all the nonzero 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
In an orthonormal coordinate system, a circle has equation x^{2}+y^{2}=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]. 
x^{2} + y^{2} = 1 y = m (x  2)The xvalues are the roots x_{1} and x_{2} of
(1 + m^{2}) x^{2}  4 m^{2} x + 4 m^{2}  1 = 0The yvalues are the roots y_{1} and y_{2} of
(1 + m^{2}) y^{2} + 4 m y + 3 m^{2} = 0 The xvalue of M is (x_{1} + x_{2})/2 = 2 m^{2}/(1 + m^{2}) The yvalue of M is (y_{1} + y_{2})/2 =  2 m /(1 + m^{2}) So, we can say that M is the intersection point of the associated lines x = 2 m^{2}/(1 + m^{2}) (1) y =  2 m /(1 + m^{2}) (2)The locus arises from eliminating m from these equations.
(1) <=> (x  2) m^{2} + x = 0 (3) (2) <=> y m^{2} + 2 m + y = 0 (4)We eliminate m.
(x  2) m^{2} + x = 0 (3) (x2) m  y = 0 (5)We now bring m from the second equation in the first equation. After simplification, we have
x^{2} + y^{2}  2x = 0 <=> (x  1)^{2} + y^{2} = 1The 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 x^{2}+y^{2}=1. Therefore, we say that all the other points of the locus are parasitic points.
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 distanceThe 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} + y^{2} = m^{2}
C' has equation (x + c)^{2} + y^{2} = (2a  m)^{2}
We find the locus when we eliminate the parameter m from the system
/ (x  c)^{2} + y^{2} = m^{2} \ (x + c)^{2} + y^{2} = 4 a^{2}  4 a m + m^{2}But that parameter m appears quadratically in each equation of the system. Using the first equation we ensure that m^{2} disappears from the second equation.
/ (x  c)^{2} + y^{2} = m^{2} \ (x + c)^{2} + y^{2} = 4 a^{2}  4 a m + (x  c)^{2} + y^{2}We calculate m from the last equation.
m = a  (c/a) xWe substitute this value in the first equation of the system.
(x  c)^{2} + y^{2} =( a  (c/a) x )^{2}This is the equation of the locus, but we'll simplify this equation.
(1 (c/a)^{2}) x^{2} + y^{2} = a^{2} c^{2} Say a^{2}  c^{2} = b^{2} <=> (b^{2}/a^{2}) x^{2} + y^{2} = b^{2} <=> x^{2}/a^{2} + y^{2}/b^{2} = 1This curve is an ellipse. F and F 'are called the foci of the ellipse.
Note that a point P on the ellipse is if and only if  PF  +  PF  = 2a.
The line d is a variable line through the fixed point A (a, 0)
and that line d cuts the yaxis in point S. The perpendicular from S on d
intersects the xaxis in point B. Find the locus of the mirror image P of point B relative to the line d. 
S(0, a m) Line BS has as equation y + a m = (1/m) x Then we find B(am^{2}, 0) and P(a m^{2}, 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 m^{2} 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 m^{2} y = 2 a mOne find a y^{2} = 4 x. That curve is a parabola with the xaxis as symmetry axis.
Then, all points of these coinciding associated lines are points of the locus.
We say that these points form the singular part of the locus.
P.Q = P.Q. cos(t)
P = Q <=> (x,y) = (x',y') <=> (P.E1 = Q.E1 en P.E2 = Q.E2)
Here is an illustration of the concept.
Choose the Xaxis on the line l.
Choose the origin of the orthonormal
coordinate system in point P, on the moment P is on the line l.
For each position of the circle, we have (vectors in bold)
P = Q + QA + AP <=> P.E1 = Q.E1 + QA.E1 + AP.E1 P.E2 = Q.E2 + QA.E2 + AP.E2 <=> x = r.t + 0 + r.cos(pi/2  t) y = 0 + r + r.cos(pi  t) <=> x = r(t  sin(t)) y = r(1  cos(t))The last equations are the parametric equations of the cycloid.
Here is an illustration of the concept.
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 cos^{3}(t)  3cos(t) sin(3t) = 3 sin(t)  4 sin^{3}(t) we find the parametric equations of the astroid from (1) and (2) x = R cos^{3}(t) y = R sin^{3}(t)The cartesian equation can be found through elimination of t.
cos(t) = (x/R)^{1/3} sin(t) = (y/R)^{1/3} => cos^{2}(t) + sin^{2}(t) = (x/R)^{2/3} + (y/R)^{2/3} <=> x^{2/3} + y^{2/3} = R^{2/3}
The parametric equations of an astroid are x= cos^{3}(t) y=sin^{3}(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(cos^{3}(t), sin^{3}(t)) on the astroid.
The slope of the tangent line in A is dy 3 sin^{2}(t) cos(t)  =  =  tan(t) dx 3 cos^{2}(t) sin(t)Now, we take a variable point B(cos^{3}(t'), sin^{3}(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  sin^{3}(t) = tan(t) ( x  cos^{3}(t)) <=> y = tan(t).x + sin(t) cos^{2}(t) + sin^{3}(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.
x = (1/2) sin(2t+pi) ( sin(t + pi/2) + cos(t + pi/2) ) y = (1/2) sin(2t+pi) ( cos(t + pi/2)  sin(t + pi/2) ) <=> x = (1/2) sin(2t) ( cos(t)  sin(t) ) y = (1/2) sin(2t) ( sin(t)  cos(t) ) <=> x = (1/2) sin(2t) ( sin(t)  cos(t) ) y = (1/2) sin(2t) ( sin(t) + cos(t) )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.
Find all points P(x,y) such that sin^{2}(x) = cos^{2}(y) 
Point A(0,1) is fixed and point P(p,0) is variable. The perpendicular to AP through point P intersects the yaxis in point M. Find the locus of the point Q such that P is the center of the segment MQ. 
A circle C_{1} with radius 3 has its center in O(0,0) and a circle C_{2}
with radius 1 has its center in D(6,0). Draw a tangent from a point P to C_{1} and C_{2}. The tangent points are A and B respectively. Find the locus of point P such that PA^{2} +PB^{2} = 58. 
A circle C_{1} with radius R=5 has its center in O(0,0) and a circle C_{2}
with radius r = 3 has its center in A(10,0). Draw a tangent from a point P to C_{1} and C_{2}. 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} = R^{2} r^{2}. 
A circle C_{1} with radius R = 5 has its center in A(5,0) and a circle C_{2}
with radius r = 3 has its center in B(3,0). A variable line d_{1} through O intersects C_{1} a second time in point D. Through O we draw the line d_{2} perpendicular to d_{1}. The line d_{2} intersects C_{2} a second time in point E. Find the locus of the center M of [DE]. Hint: Sometimes geometry is stronger than 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. 
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. 
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. 
On the circle with equation x^{2} + (y1)^{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. 