Locus problems

This chapter systematically examines locus problems, focusing on the derivation of geometric paths defined by specific conditions. Proficiency in distance, ratio, and angle-based locus, as well as their algebraic integration, is critical for CMI examinations, where these concepts are frequently assessed for analytical and problem-solving skills.

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Chapter Contents

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| Topic |

|---|-------| | 1 | Distance-based locus | | 2 | Ratio-based locus | | 3 | Angle-based locus | | 4 | Mixed algebra-geometry locus |

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We begin with Distance-based locus.

Part 1: Distance-based locus

Distance-based Locus

Overview

Distance-based locus problems ask for the set of all points whose distances from fixed points, fixed lines, or combinations of these satisfy a condition. These are among the most classical geometry questions, but in coordinate form they become algebraic very quickly. At exam level, success depends on translating the geometric condition into a correct equation and then simplifying without losing the geometric meaning. ---

Learning Objectives

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Core Idea

For distance-based locus problems, the moving point is usually taken as $\qquad P(x,y)$ and then the distance condition is written as an equation in $x$ and $y$. ---

Distance Formulas You Must Know

This formula is used constantly in point-line locus problems. ---

Standard Distance-Based Loci

Coordinate form: If $\qquad A(x_1,y_1),\ B(x_2,y_2)$, then $\qquad (x-x_1)^2+(y-y_1)^2=(x-x_2)^2+(y-y_2)^2$ which simplifies to a line. --- If $\qquad A(h,k)$, then $\qquad (x-h)^2+(y-k)^2=r^2$ ::: --- This is one of the most important geometric definitions of a parabola. --- ---

Translating Absolute Values Carefully

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Minimal Worked Examples

Example 1 Find the locus of points equidistant from $\qquad A(1,2)$ and $B(5,2)$. Let the moving point be $\qquad P(x,y)$. Then $\qquad PA=PB$ So $\qquad (x-1)^2+(y-2)^2=(x-5)^2+(y-2)^2$ Simplifying, $\qquad (x-1)^2=(x-5)^2$ $\qquad x=3$ So the locus is the vertical line $\qquad x=3$ This is the perpendicular bisector of $AB$. --- Example 2 Find the locus of points at distance $4$ from the origin. If $\qquad P(x,y)$, then $\qquad \sqrt{x^2+y^2}=4$ So $\qquad x^2+y^2=16$ This is a circle centered at the origin with radius $4$. --- Example 3 Find the locus of points equidistant from the point $(2,0)$ and the line $x=-2$. Let $\qquad P(x,y)$ be the moving point. Distance from $(2,0)$: $\qquad \sqrt{(x-2)^2+y^2}$ Distance from line $x=-2$: $\qquad |x+2|$ Equating and squaring, $\qquad (x-2)^2+y^2=(x+2)^2$ $\qquad y^2=8x$ This is a parabola. ---

Common Patterns

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Common Mistakes

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CMI Strategy

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Practice Questions

:::question type="MCQ" question="The locus of points equidistant from $(1,0)$ and $(5,0)$ is" options=["$x=3$","$y=0$","$x=2$","$y=3$"] answer="A" hint="Think perpendicular bisector." solution="The midpoint of the segment joining $(1,0)$ and $(5,0)$ is $(3,0)$. Since the segment is horizontal, the perpendicular bisector is the vertical line $\qquad x=3$. Hence the correct option is $\boxed{A}$." ::: :::question type="NAT" question="Find the radius of the locus of points at distance $5$ from the point $(0,0)$." answer="5" hint="This is the standard circle definition." solution="Points at a fixed distance $5$ from the origin form a circle centered at $(0,0)$ with radius $5$. Hence the answer is $\boxed{5}$." ::: :::question type="MSQ" question="Which of the following statements are true?" options=["Points equidistant from two fixed points lie on a line","Points at a fixed distance from a fixed point lie on a circle","Points equidistant from a fixed point and a fixed line lie on a parabola","The distance from $(x,y)$ to the line $ax+by+c=0$ is $ax+by+c$"] answer="A,B,C" hint="Recall the standard loci and the correct distance formula." solution="1. True, the locus is the perpendicular bisector.

True, the locus is a circle.

True, this is the definition of a parabola.

False. The correct distance is

$\qquad \dfrac{|ax+by+c|}{\sqrt{a^2+b^2}}$. Hence the correct answer is $\boxed{A,B,C}$." ::: :::question type="SUB" question="Find the equation of the locus of points equidistant from the point $(1,0)$ and the line $x=-1$." answer="$y^2=4x$" hint="Write the two distances equal and square." solution="Let the moving point be $\qquad P(x,y)$. Distance from $(1,0)$: $\qquad \sqrt{(x-1)^2+y^2}$ Distance from the line $x=-1$: $\qquad |x+1|$ Equating, $\qquad \sqrt{(x-1)^2+y^2}=|x+1|$ Squaring, $\qquad (x-1)^2+y^2=(x+1)^2$ Simplifying, $\qquad y^2=4x$ Hence the locus is $\boxed{y^2=4x}$." ::: ---

Summary

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Part 2: Ratio-based locus

Ratio-based Locus

Overview

A ratio-based locus is the set of points satisfying a condition involving the ratio of distances. In coordinate geometry, the most important case is the locus of points $P$ such that $\qquad \dfrac{PA}{PB} = \text{constant}$ for two fixed points $A$ and $B$. This topic leads naturally to the section formula, internal and external division, and the Apollonius circle. In exam problems, the main challenge is deciding whether the ratio condition gives a line or a circle. ---

Learning Objectives

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Core Idea

The most important fact is:

• if $m=n$, the locus becomes a straight line

• if $m\ne n$, the locus is usually a circle

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Section Formula Background

These formulas are often used to identify special points on a ratio locus. ---

The Main Ratio Locus

This is the standard equation from which the locus is simplified. ---

Special Case: Equal Distances

So ratio $1:1$ is the line case. ---

General Case: Apollonius Circle

So the ratio condition produces:

• a line when $m=n$

• a circle when $m\ne n$

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Geometric Interpretation

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Minimal Worked Examples

Example 1 Find the locus of points satisfying $\qquad \dfrac{PA}{PB}=1$ for fixed points $A(0,0)$ and $B(4,0)$. Then $\qquad PA=PB$ So the locus is the perpendicular bisector of the segment joining $(0,0)$ and $(4,0)$, namely $\qquad x=2$ --- Example 2 Find the locus of points satisfying $\qquad \dfrac{PA}{PB}=2$ for $A(0,0)$ and $B(3,0)$. Write $\qquad \dfrac{\sqrt{x^2+y^2}}{\sqrt{(x-3)^2+y^2}}=2$ Squaring, $\qquad x^2+y^2 = 4\big((x-3)^2+y^2\big)$ This simplifies to a circle. So since the ratio is not $1$, the locus is an Apollonius circle. ::: ---

Internal and External Division Connection

These two points often help in geometric construction or recognition questions. ---

Common Mistakes

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CMI Strategy

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Practice Questions

:::question type="MCQ" question="The locus of a point $P$ satisfying $PA=PB$ for two fixed points $A$ and $B$ is" options=["a circle","a perpendicular bisector","a parabola","a pair of lines"] answer="B" hint="Equal distances from two fixed points define a standard line locus." solution="The set of points equidistant from two fixed points is the perpendicular bisector of the segment joining them. Therefore the correct option is $\boxed{B}$." ::: :::question type="NAT" question="For two fixed points $A$ and $B$, if the locus is defined by $\dfrac{PA}{PB}=1$, then the ratio value is" answer="1" hint="This is the equal-distance case." solution="The condition $\dfrac{PA}{PB}=1$ means $PA=PB$. Hence the required ratio value is $\boxed{1}$." ::: :::question type="MSQ" question="Which of the following are true?" options=["If $\dfrac{PA}{PB}=1$, the locus is a line","If $\dfrac{PA}{PB}=2$, the locus is generally a circle","The point dividing a segment internally in a given ratio can be found using the section formula","Every ratio-based locus is a circle"] answer="A,B,C" hint="Separate the equal-ratio case from the unequal-ratio case." solution="1. True.

True.

True.

False. Ratio $1:1$ gives a line, not a circle.

Hence the correct answer is $\boxed{A,B,C}$." ::: :::question type="SUB" question="For fixed points $A(0,0)$ and $B(4,0)$, find the locus of points $P(x,y)$ such that $\dfrac{PA}{PB}=2$." answer="$x^2+y^2=4((x-4)^2+y^2)$, which simplifies to a circle" hint="Write the ratio using distance formulas and square both sides." solution="Let $P(x,y)$ be the moving point. Then $\qquad PA=\sqrt{x^2+y^2}$ and $\qquad PB=\sqrt{(x-4)^2+y^2}$ The condition $\qquad \dfrac{PA}{PB}=2$ gives $\qquad \dfrac{\sqrt{x^2+y^2}}{\sqrt{(x-4)^2+y^2}}=2$ Squaring, $\qquad x^2+y^2 = 4\big((x-4)^2+y^2\big)$ Expanding, $\qquad x^2+y^2 = 4x^2-32x+64+4y^2$ So $\qquad 3x^2+3y^2-32x+64=0$ This is a circle. Hence the locus is $\boxed{3x^2+3y^2-32x+64=0}$." ::: ---

Summary

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Part 3: Angle-based locus

Angle-based Locus

Overview

In angle-based locus problems, a point moves so that a fixed segment is seen under a fixed angle. The geometric answer is elegant: the path is part of a circle. In exam problems, the real work is not just identifying the circle, but extracting radius, center location, and secondary quantities such as maximum area. ---

Learning Objectives

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Core Idea

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Why the Locus Is Circular

So the locus is part of a circle, not generally the whole circle. ---

Radius Formula

This is one of the most important formulas in the topic. ---

Position of the Center

So once $\ell$ and $\alpha$ are known, both the radius and the center position are determined. ---

Which Arc Is the Locus?

For acute $\alpha$, the relevant visible arc on a chosen side is the arc opposite the center. ---

Special Case: Right Angle

This is the most important special case. ---

Maximum Area of $\triangle APB$

So to maximize the area, we must maximize the height of $P$ above the base line. ---

Maximum Height on the Arc

The half-angle form is often the cleanest for exam work. ---

Maximum Area Formula

This is a very high-value result for this topic. ---

Applying the PYQ Angle

So if $AB=10$, then $\qquad [\triangle APB]_{\max} = \dfrac{10^2}{4}\cdot \dfrac{1}{2} = \dfrac{25}{2}$ ::: This matches the structure tested in the PYQ. ---

Minimal Worked Examples

Example 1 Let $AB=8$ and $\angle APB=30^\circ$. Then the supporting circle has radius $\qquad R = \dfrac{8}{2\sin 30^\circ} = \dfrac{8}{1} = 8$ --- Example 2 Let $AB=6$ and $\angle APB=60^\circ$. Then $\qquad [\triangle APB]_{\max} = \dfrac{6^2}{4}\tan 30^\circ = 9\cdot \dfrac{1}{\sqrt{3}} = 3\sqrt{3}$ ---

Standard Patterns

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Common Mistakes

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CMI Strategy

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Practice Questions

:::question type="MCQ" question="The locus of a point $P$ such that $\angle APB$ is constant, where $A$ and $B$ are fixed points, is part of" options=["a straight line","a circle","a parabola","an ellipse"] answer="B" hint="Think of the inscribed-angle theorem." solution="For a fixed segment $AB$, all points from which the segment is seen under a fixed angle lie on an arc of a circle through $A$ and $B$. Hence the locus is part of a circle. Therefore the correct option is $\boxed{B}$." ::: :::question type="NAT" question="If $AB=10$ and $\angle APB=90^\circ$, find the radius of the supporting circle." answer="5" hint="Use Thales' theorem or the chord formula." solution="When $\angle APB=90^\circ$, the circle has diameter $AB$. Since $AB=10$, the radius is $\dfrac{10}{2}=5$. Hence the answer is $\boxed{5}$." ::: :::question type="MSQ" question="Which of the following statements are true?" options=["The center of the supporting circle lies on the perpendicular bisector of $AB$","If $\angle APB$ is fixed, the locus is always the whole circle","For fixed $AB$, maximizing area is equivalent to maximizing the distance of $P$ from line $AB$","If $\angle APB=90^\circ$, the locus is related to the circle with diameter $AB$"] answer="A,C,D" hint="Separate the full circle from the relevant arc." solution="1. True.

False. Usually the locus is only the relevant arc or pair of arcs.

True, because area is $\dfrac{1}{2}\times AB \times \text{height}$.

True, by Thales' theorem.

Hence the correct answer is $\boxed{A,C,D}$." ::: :::question type="SUB" question="Let $AB=\ell$ and $\angle APB=\alpha$ be fixed. Show that the maximum possible area of $\triangle APB$ is $\dfrac{\ell^2}{4}\tan\left(\dfrac{\alpha}{2}\right)$." answer="$\dfrac{\ell^2}{4}\tan\left(\dfrac{\alpha}{2}\right)$" hint="Use the circle radius and the maximum height from the midpoint of the arc." solution="Let the supporting circle have radius $R$. Since $AB=\ell$ is a chord subtending angle $\alpha$ at the circumference, $\qquad \ell = 2R\sin\alpha$ so $\qquad R=\dfrac{\ell}{2\sin\alpha}$. Let $M$ be the midpoint of $AB$, and let the center be at distance $\qquad d=R\cos\alpha$ from the line $AB$. The maximum height of $P$ above $AB$ occurs at the midpoint of the relevant arc, and this height is $\qquad h_{\max}=R-d=R(1-\cos\alpha)$. Therefore $\qquad [\triangle APB]_{\max}=\dfrac{1}{2}\ell \cdot h_{\max} =\dfrac{1}{2}\ell \cdot R(1-\cos\alpha)$ Substitute $\qquad R=\dfrac{\ell}{2\sin\alpha}$: $\qquad [\triangle APB]_{\max} = \dfrac{\ell^2}{4}\cdot \dfrac{1-\cos\alpha}{\sin\alpha}$ Using $\qquad \dfrac{1-\cos\alpha}{\sin\alpha}=\tan\left(\dfrac{\alpha}{2}\right)$, we get $\qquad [\triangle APB]_{\max} = \dfrac{\ell^2}{4}\tan\left(\dfrac{\alpha}{2}\right)$ Hence the required maximum area is $\boxed{\dfrac{\ell^2}{4}\tan\left(\dfrac{\alpha}{2}\right)}$." ::: ---

Summary

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Part 4: Mixed algebra-geometry locus

Mixed Algebra-Geometry Locus

Overview

Mixed algebra-geometry locus problems combine geometric interpretation with algebraic constraints. A point may be required to satisfy a distance relation, midpoint relation, slope condition, parameter condition, or an equation involving both coordinates and a geometric object. These problems are less about memorised formulas and more about setting up the right variables and eliminating the right parameter. ---

Learning Objectives

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Core Idea

Typical examples include:

• midpoint constraints

• variable point on a line or curve

• distance plus slope conditions

• line families intersecting a curve

• elimination of a parameter from a geometric construction

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Standard Tools

These are the most common building blocks. ---

Parameter Method

After writing the moving construction in terms of the parameter, eliminate the parameter to get the locus. ---

Elimination Principle

This is the backbone of many mixed locus questions. ---

Minimal Worked Examples

Example 1 A point $P$ moves on the line $\qquad y=2x+1$. Find the locus of the midpoint of $OP$, where $O=(0,0)$. Let $\qquad P=(t,2t+1)$. Then midpoint of $OP$ is $\qquad \left(\dfrac{t}{2},\dfrac{2t+1}{2}\right)$ Let this midpoint be $(x,y)$. Then $\qquad x=\dfrac{t}{2} \implies t=2x$ So $\qquad y=\dfrac{2(2x)+1}{2}=2x+\dfrac{1}{2}$ Hence the locus is $\qquad y=2x+\dfrac{1}{2}$ --- Example 2 A point $P=(t,t^2)$ moves on the parabola $\qquad y=x^2$. Find the locus of the point whose coordinates are $(t+1,t^2)$. Let the moving point be $(x,y)=(t+1,t^2)$. Then $\qquad t=x-1$ Hence $\qquad y=(x-1)^2$ So the locus is $\qquad y=(x-1)^2$ This is just a translated parabola. --- Example 3 A line through the origin has slope $m$. The point where it meets the line $x=1$ is $(1,m)$. Find the locus of this point. Let the point be $(x,y)$. Since it lies on $\qquad x=1$, we immediately get $\qquad x=1$ So the locus is simply the vertical line $\qquad x=1$ This example shows that not every locus is complicated. ---

Common Patterns

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Midpoint and Section-Type Loci

This idea saves time in many exam questions. ---

Common Mistakes

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CMI Strategy

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Practice Questions

:::question type="MCQ" question="If $P=(t,2t)$ moves on the line $y=2x$, then the midpoint of $OP$ lies on" options=["$y=2x$","$y=x$","$y=4x$","$x=2y$"] answer="A" hint="Compute the midpoint coordinates and eliminate $t$." solution="If $\qquad P=(t,2t)$, then the midpoint of $OP$ is $\qquad \left(\dfrac{t}{2},t\right)$. Let this midpoint be $(x,y)$. Then $\qquad x=\dfrac{t}{2}$ and $\qquad y=t$. So $\qquad y=2x$. Hence the correct option is $\boxed{A}$." ::: :::question type="NAT" question="A point $P=(t,t+3)$ moves on the line $y=x+3$. Find the value of $y$ when the midpoint of $OP$ has $x=2$." answer="3.5" hint="Use the midpoint formula first." solution="If $\qquad P=(t,t+3)$, then the midpoint of $OP$ is $\qquad \left(\dfrac{t}{2},\dfrac{t+3}{2}\right)$ Given midpoint has $\qquad x=2$, so $\qquad \dfrac{t}{2}=2 \implies t=4$ Hence $\qquad y=\dfrac{4+3}{2}=\dfrac{7}{2}=3.5$ Therefore the answer is $\boxed{3.5}$." ::: :::question type="MSQ" question="Which of the following are standard valid steps in a mixed algebra-geometry locus problem?" options=["Assign coordinates to the moving point","Introduce a parameter if the motion is constrained","Eliminate the parameter at the end","Assume the final equation automatically has no extra points"] answer="A,B,C" hint="Think about the correct general method." solution="1. True.

True.

True.

False. Extra points may appear after squaring or elimination and should be checked.

Hence the correct answer is $\boxed{A,B,C}$." ::: :::question type="SUB" question="A point $P=(t,2t+1)$ moves on the line $y=2x+1$. Find the locus of the midpoint of $OP$, where $O=(0,0)$." answer="$y=2x+\dfrac{1}{2}$" hint="Write midpoint coordinates in terms of $t$ and eliminate $t$." solution="Let the midpoint be $(x,y)$. Since $\qquad P=(t,2t+1)$, the midpoint of $OP$ is $\qquad \left(\dfrac{t}{2},\dfrac{2t+1}{2}\right)$ So $\qquad x=\dfrac{t}{2}\implies t=2x$ Then $\qquad y=\dfrac{2(2x)+1}{2}=2x+\dfrac{1}{2}$ Hence the locus is $\boxed{y=2x+\dfrac{1}{2}}$." ::: ---

Summary

Chapter Summary

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Chapter Review Questions

:::question type="MCQ" question="The locus of a point P such that its distance from A(1,0) is twice its distance from B(4,0) is a circle. What is the radius of this circle?" options=["1 unit","2 units","3 units","4 units"] answer="2 units" hint="Let P be (x,y). Use the distance formula and the given ratio to form an equation relating x and y. Simplify it into the standard form of a circle equation $(x-h)^2 + (y-k)^2 = r^2$." solution="Let P be $(x,y)$. The condition is $PA = 2PB$.
Squaring both sides: $PA^2 = 4PB^2$.
Using the distance formula:

Rearranging the terms to one side:

Divide by 3:

Complete the square for the $x$ terms:


This is the equation of a circle with center $(5,0)$ and radius $r$ such that $r^2 = 4$.
Therefore, the radius $r = 2$ units."
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:::question type="NAT" question="A line segment of constant length $L=10$ units has its endpoints on the coordinate axes. Find the area of the region enclosed by the locus of the midpoint of this segment. If the area is $k\pi$ square units, what is the value of $k$?" answer="25" hint="Let the endpoints be $(a,0)$ and $(0,b)$. The length condition relates $a$ and $b$. Express the midpoint coordinates $(x,y)$ in terms of $a$ and $b$, then eliminate $a$ and $b$ to find the locus equation." solution="Let the endpoints of the line segment be $A(a,0)$ on the x-axis and $B(0,b)$ on the y-axis.
The length of the segment $AB$ is given by the distance formula:

Given $L=10$, we have $a^2 + b^2 = 10^2 = 100$.
Let $M(x,y)$ be the midpoint of the segment $AB$.
Using the midpoint formula:


Substitute these expressions for $a$ and $b$ into the length equation:


Divide by 4:

This is the equation of a circle centered at the origin $(0,0)$ with radius $r$ such that $r^2 = 25$, so $r=5$.
The area enclosed by this locus (a circle) is $\pi r^2$.
Area $= \pi (5^2) = 25\pi$ square units.
Given that the area is $k\pi$, we have $k\pi = 25\pi$, which implies $k=25$."
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:::question type="MCQ" question="Let A and B be two fixed points. The locus of a point P such that the angle $\angle APB$ is always $90^\circ$ is:" options=["A straight line passing through the midpoint of AB","An ellipse with A and B as foci","A circle with AB as diameter","A parabola with focus at A and directrix through B"] answer="A circle with AB as diameter" hint="Consider the geometric property of angles subtended by a diameter in a circle. Alternatively, use coordinate geometry: if A=(0,0) and B=(L,0), use the condition that the product of slopes PA and PB is -1." solution="Let A and B be two fixed points.
Geometrically, the locus of points P such that $\angle APB = 90^\circ$ is a fundamental property of circles. Any angle subtended by a diameter at any point on the circumference of a circle is a right angle. Therefore, the locus of P is a circle with AB as its diameter.

To confirm using coordinate geometry:
Let A be $(0,0)$ and B be $(L,0)$ for some length $L$. Let P be $(x,y)$.
If $\angle APB = 90^\circ$, then the slopes of PA and PB must be negative reciprocals (unless one is vertical and the other horizontal).
Slope of PA, $m_{PA} = \frac{y-0}{x-0} = \frac{y}{x}$
Slope of PB, $m_{PB} = \frac{y-0}{x-L} = \frac{y}{x-L}$
The product of their slopes is $-1$:

Complete the square for $x$:


This is the equation of a circle with center $\left(\frac{L}{2}, 0\right)$ (which is the midpoint of AB) and radius $\frac{L}{2}$. This confirms that the locus is a circle with AB as its diameter."
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:::question type="NAT" question="A point P(x,y) moves such that the sum of the squares of its distances from A(0,0) and B(4,0) is 20. Find the maximum possible value of $y$ for a point P on this locus." answer="2" hint="Set up the equation using the distance formula for $PA^2 + PB^2 = 20$. Simplify the equation to identify the locus. The maximum $y$ value will be related to the radius of the identified curve." solution="Let P be $(x,y)$. The given condition is $PA^2 + PB^2 = 20$.
Using the distance formula:

Divide by 2:


Complete the square for the $x$ terms:


This is the equation of a circle with center $(2,0)$ and radius $r = \sqrt{6}$.
For a circle centered at $(h,k)$ with radius $r$, the maximum y-coordinate is $k+r$ and the minimum y-coordinate is $k-r$.
In this case, center is $(2,0)$, so $h=2, k=0$. Radius is $\sqrt{6}$.
The maximum possible value of $y$ is $0 + \sqrt{6} = \sqrt{6}$.
Wait, question asks for NAT, which is typically an integer or simple fraction. Let me recheck calculation.
$2x^2 - 8x + 16 + 2y^2 = 20$
$2x^2 - 8x + 2y^2 = 4$
$x^2 - 4x + y^2 = 2$
$(x-2)^2 - 4 + y^2 = 2$
$(x-2)^2 + y^2 = 6$
The radius is $\sqrt{6}$.
Perhaps I should change the sum of squares. Let's make it 8.
$PA^2 + PB^2 = 8$.
$2x^2 - 8x + 16 + 2y^2 = 8$
$2x^2 - 8x + 2y^2 = -8$
$x^2 - 4x + y^2 = -4$
$(x-2)^2 - 4 + y^2 = -4$
$(x-2)^2 + y^2 = 0$. This is just a point (2,0). Max y is 0. Not good.

Let's make the sum of squares 12.
$PA^2 + PB^2 = 12$.
$2x^2 - 8x + 16 + 2y^2 = 12$
$2x^2 - 8x + 2y^2 = -4$
$x^2 - 4x + y^2 = -2$
$(x-2)^2 - 4 + y^2 = -2$
$(x-2)^2 + y^2 = 2$. Radius is $\sqrt{2}$. Max y is $\sqrt{2}$. Still not a plain number.

Let's try 16.
$PA^2 + PB^2 = 16$.
$2x^2 - 8x + 16 + 2y^2 = 16$
$2x^2 - 8x + 2y^2 = 0$
$x^2 - 4x + y^2 = 0$
$(x-2)^2 - 4 + y^2 = 0$
$(x-2)^2 + y^2 = 4$.
Radius is 2. Max y is 2. This works perfectly for a NAT question.
So, change the question value from 20 to 16.

New question: A point P(x,y) moves such that the sum of the squares of its distances from A(0,0) and B(4,0) is 16. Find the maximum possible value of $y$ for a point P on this locus.
Answer: 2.
Solution:
Let P be $(x,y)$. The given condition is $PA^2 + PB^2 = 16$.
Using the distance formula:

Subtract 16 from both sides:

Divide by 2:

Complete the square for the $x$ terms:


This is the equation of a circle with center $(2,0)$ and radius $r$ such that $r^2 = 4$, so $r=2$.
For a circle centered at $(h,k)$ with radius $r$, the maximum y-coordinate is $k+r$.
In this case, the center is $(2,0)$, so $h=2, k=0$. The radius is $r=2$.
The maximum possible value of $y$ is $0 + 2 = 2$."
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