Lesson Derive an equation and show that the locus of points is a circle
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<H2>Derive an equation and show that the locus of points is a circle</H2> <H3>Problem 1</H3>Find the locus of points P(x,y) such that the distance from P to point (3,0) is twice its distance to point (1,0). <B>Solution</B> <pre> Let (x,y) be the point of this locus. Then {{{sqrt((x-3)^2 + (y-0)^2)}}} = {{{2*sqrt((x-1)^2 + (y-0)^2)}}}. (1) The left side is the distance from (x,y) to point (3,0). The right side is the doubled distance from (x,y) to point (1,0). Now square both sides of (1). You will get {{{(x-3)^2 + y^2}}} = {{{4*(x-1)^2 + 4*y^2}}}. Simplify: {{{x^2 - 6x + 9 + y^2}}} = {{{4x^2 - 8x + 4 + 4y^2}}}, {{{3x^2 + 3y^2 - 2x}}} = {{{5}}}, {{{(x^2-(2/3)x) + y^2}}} = {{{5/3}}}, {{{(x^2-(2/3)x + (1/3)^2) + y^2}}} = {{{5/3 + (1/3)^2}}}, {{{(x-1/3)^2 + y^2}}} = {{{16/9}}}, {{{(x-1/3)^2 + y^2}}} = {{{(4/3)^2}}}. It is the standard equation of the circle of the radius {{{4/3}}} with the center at the point ({{{1/3}}},{{{0}}}). <U>Answer</U>. The locus of points under the problem's question is the circle of the radius {{{4/3}}} centered at the point ({{{1/3}}},{{{0}}}). {{{graph( 330, 330, -2.5, 4.5, -3.5, 3.5, sqrt((4/3)^2-(x-1/3)^2), -sqrt((4/3)^2-(x-1/3)^2) )}}} Circle {{{(x-1/3)^2 + y^2}}} = {{{(4/3)^2}}} </pre> <H3>Problem 2</H3>If {{{(z-1)/(z+1)}}} is purely imaginary, then what is the locus of points z ? <B>Solution</B> <pre> The points (1,0) and (-1,0) are selected points in the complex number plane. For an arbitrary complex number z, the numbers z-1 and z+1 are vectors, connecting z with the points 1 and -1. The condition that {{{(z-1)/(z+1)}}} is purely imaginary means that the vectors z-1 and z+1 are perpendicular. So, the problem asks to find points z in complex plane, such that the "visibilty angle" from z to these points "1" and "-1" is the right angle. Or, in other terms, find points z in complex plane, such that the angle between the vectors z-1 and z+1 is the right angle. Clearly, these points are on the unit circle, and they provide the right angle leaning on the segment [-1,1] as on the diameter of the unit circle. Thus the sought locus is the unit circle. The endpoints (-1,0) and (1,0) are not included to the locus. </pre> <H3>Problem 3</H3>If the foci of ellipse {{{x^2/9}}} + {{{y^2/4 = 1}}} subtend a right angle at some point P, find the locus of points P. <B>Solution</B> <pre> The foci of the ellipse {{{x^2/9}}} + {{{y^2/4}}} = 1 are the points (-c,0) and (c,0) on x-axis, where c = {{{sqrt(9-4)}}} = {{{sqrt(5)}}}. The locus of points P such that the foci subtend the right angle at point P is the circle, centered at the origin of the coordinate plane and having the radius of {{{sqrt(5)}}}, EXCLUDING the endpoints of its diameter that lie on x-axis. </pre> <H3>Problem 4</H3>Find the locus of points representing complex numbers z such that arg{{{(z - 2)/(z + 2)}}} = {{{pi/3}}}. <B>Solution</B> <pre> In this problem we have two selected points in the complex plane: A = (-2,0) and B = (2,0). The complex number z-2 represents a vector connecting point z with the point (2,0). The complex number z+2 represents a vector connecting point z with the point (-2,0). Since {{{arg((z-2)/(x+2))}}} = {{{pi/3}}} (given), it means that the angle between the vectors (z-2) and (z+2) is {{{pi/3}}}. It means that the point z lies on the arc of a circle and the vectors (z-2) and (z+2) form an inscribed angle of the measure of {{{pi/3}}} = 60 degrees, which leans on the segment AB as on the chord. The locus is shown in the Figure below in red line. The parts of the circles shown in black are not the parts of the locus. {{{drawing( 400, 300, -4, 4, -2, 4, line( -3.5, 0, 3.5, 0 ), blue(line( -2, 0.01, 2, 0.01 )), blue(line( -2, -0.03, 2, -0.03 )), locate(-2.4, 0.03, "-2") locate( 2.1, 0.03, "2") line( -2, 0, 2, 0 ), circle (0, 2*sqrt(3)/3, 4/sqrt(3), 4/sqrt(3)), circle (0, 2*sqrt(3)/3, 0.05, 0.05), locate (0.2, 2*sqrt(3)/3+0.2, P), red(arc (0, 2*sqrt(3)/3, 2*(4/sqrt(3)+0.03), 2*(4/sqrt(3)+0.03), 150, 390)), red(arc (0, 2*sqrt(3)/3, 2*(4/sqrt(3)+0.00), 2*(4/sqrt(3)+0.00), 150, 390)), red(arc (0, 2*sqrt(3)/3, 2*(4/sqrt(3)-0.03), 2*(4/sqrt(3)-0.03), 150, 390)), line(-2, 0, -1.1547, 3.1547), line( 2, 0, -1.1547, 3.1547), locate (-1.5, 1.8, "z+2"), locate ( 0.5, 1.8, "z-2") )}}} </pre> My other lessons in this site on deducing equations for locuses of points are - <A HREF=https://www.algebra.com/algebra/homework/Quadratic-relations-and-conic-sections/Derive-an-equation-that-a-locus-of-points-is-a-STRAIGHT-LINE.lesson>Derive an equation and show that the locus of points is a straight line</A> - <A HREF=https://www.algebra.com/algebra/homework/Quadratic-relations-and-conic-sections/Derive-an-equation-and-show-that-the-locus-of-points-is-an-ELLIPSE.lesson>Derive an equation and show that the locus of points is an ellipse</A> - <A HREF=https://www.algebra.com/algebra/homework/Quadratic-relations-and-conic-sections/Derive-an-equation-and-show-that-the-locus-of-points-is-a-HYPERBOLA.lesson>Derive an equation and show that the locus of points is a hyperbola</A> - <A HREF=https://www.algebra.com/algebra/homework/Quadratic-relations-and-conic-sections/Derive-an-equation-and-show-that-the-locus-of-points-is-a-PARABOLA.lesson>Derive an equation and show that the locus of points is a parabola</A> - <A HREF=https://www.algebra.com/algebra/homework/Quadratic-relations-and-conic-sections/OVERWIEW-of-lessons-on-deducing-equations--for-locuses-of-points.lesson>OVERWIEW of lessons on deducing equations for locuses of points</A> Use this file/link <A HREF=https://www.algebra.com/algebra/homework/complex/ALGEBRA-II-YOUR-ONLINE-TEXTBOOK.lesson>ALGEBRA-II - YOUR ONLINE TEXTBOOK</A> to navigate over all topics and lessons of the online textbook ALGEBRA-II.
