Why do we square both sides? Doesn't that introduce extraneous solutions?

Squaring is a standard technique to eliminate radicals. Since distance is always non-negative, $PA = 3PB$ is equivalent to $PA^2 = (3PB)^2$ for all points $P$. There are no extraneous solutions here because both sides are inherently positive.

How do I know it's a circle without expanding everything?

This is a specific case of the Circle of Apollonius. Whenever the ratio of distances from two fixed points is constant (and not equal to 1), the locus is always a circle.

Why is the final answer 56 and not 56.25?

The question asks for the nearest integer. Since 56.25 is closer to 56 than to 57, we round down.

JEE Main 2021 (24 February Shift 2)

LEVELJEE Main

Animated Solution for Mathematics - Circles: Let a point $P$ be such that its distance from the point $(5, 0)$ is thrice the distance of $P$ from the point $(- 5, 0)$ . If the locus of the point $P$ is a circle of radius $r$ , then $4 r^{2}$ is equal to (Round off to the nearest integer)

Enter Numerical Value:

Visualized Solution

$A (5, 0)$ and $B (- 5, 0)$

Let the fixed points be $A (5, 0)$ and $B (- 5, 0)$ .
Let the moving point be $P (x, y)$ .

$P A = 3 P B$

Given condition: Distance of $P$ from $(5, 0)$ is thrice its distance from $(- 5, 0)$ .
Mathematically: $P A = 3 P B$

$P A^{2} = 9 P B^{2}$

To remove square roots, square both sides:
$P A^{2} = (3 P B)^{2}$
$P A^{2} = 9 P B^{2}$

$d^{2} = (x_{2} - x_{1})^{2} + (y_{2} - y_{1})^{2}$

Using distance formula $d^{2} = (x_{2} - x_{1})^{2} + (y_{2} - y_{1})^{2}$ :
$(x - 5)^{2} + (y - 0)^{2} = 9 [(x - (- 5))^{2} + (y - 0)^{2}]$

$(x - 5)^{2} + y^{2}$

Expand LHS using $(a - b)^{2} = a^{2} - 2 ab + b^{2}$ :
$x^{2} - 10 x + 25 + y^{2} = 9 [(x + 5)^{2} + y^{2}]$

$9 [(x + 5)^{2} + y^{2}]$

Expand RHS and distribute the constant $9$ :
$x^{2} - 10 x + 25 + y^{2} = 9 (x^{2} + 10 x + 25 + y^{2})$

$8 x^{2} + 8 y^{2} + 100 x + 200 = 0$

Rearrange all terms to one side:
$9 x^{2} - x^{2} + 9 y^{2} - y^{2} + 90 x + 10 x + 225 - 25 = 0$
$8 x^{2} + 8 y^{2} + 100 x + 200 = 0$

$x^{2} + y^{2} + \frac{25}{2} x + 25 = 0$

Divide the entire equation by $8$ :
$x^{2} + y^{2} + \frac{100}{8} x + \frac{200}{8} = 0$
$x^{2} + y^{2} + \frac{25}{2} x + 25 = 0$

$g, f, c$

Compare with $x^{2} + y^{2} + 2 g x + 2 f y + c = 0$ :
$2 g = \frac{25}{2} ⟹ g = \frac{25}{4}$
$2 f = 0 ⟹ f = 0$
$c = 25$

$r^{2} = g^{2} + f^{2} - c$

Radius formula: $r = g^{2} + f^{2} - c$
$r^{2} = g^{2} + f^{2} - c$
$r^{2} = (\frac{25}{4})^{2} + 0^{2} - 25$
$r^{2} = \frac{625}{16} - 25$

$r^{2} = \frac{225}{16}$

Take LCM to simplify:
$r^{2} = \frac{625 - 400}{16}$
$r^{2} = \frac{225}{16}$

$4 r^{2}$

Calculate $4 r^{2}$ :
$4 r^{2} = 4 \times \frac{225}{16}$
$4 r^{2} = \frac{225}{4} = 56.25$

$56.25 \approx 56$

Round off to the nearest integer:
$56.25 \approx 56$
Final Answer: 56

00:00 / 00:00

The Sigma Insight: Standard and General Equation of a Circle

Solution Diagram

Analyzing the Setup

Imagine you are standing on a coordinate plane with two fixed anchors: point

A

(5, 0)

and point

B

(- 5, 0)

We are tasked with finding the path of a point

P (x, y)

that maintains a specific relationship: its distance from

A

is exactly three times its distance from

B

. This is the classic Circle of Apollonius.

The Algebraic Foundation

We start with the condition

P A = 3 P B

. To eliminate the cumbersome square roots inherent in the distance formula, we square both sides to obtain:

P A^{2} = 9 P B^{2}

By substituting the coordinates of

A (5, 0)

and

B (- 5, 0)

, we derive the following equation:

(x - 5)^{2} + y^{2} = 9 ((x + 5)^{2} + y^{2})

Expanding both sides, we get:

x^{2} - 10 x + 25 + y^{2} = 9 (x^{2} + 10 x + 25 + y^{2})

The Simplification

Next, we expand the right side of the equation:

x^{2} - 10 x + 25 + y^{2} = 9 x^{2} + 90 x + 225 + 9 y^{2}

Bringing all terms to one side, we arrive at:

8 x^{2} + 8 y^{2} + 100 x + 200 = 0

Dividing the entire equation by

8

, we obtain the standard form:

x^{2} + y^{2} + 12.5 x + 25 = 0

Final Calculation

Comparing this to the general circle equation

x^{2} + y^{2} + 2 g x + 2 f y + c = 0

, we identify the parameters:

g = 6.25

f = 0

, and

c = 25

The radius squared is determined by the formula

r^{2} = g^{2} + f^{2} - c

. Substituting our values:

r^{2} = (6.25)^{2} + 0^{2} - 25

r^{2} = 39.0625 - 25 = 14.0625

The problem requires the value of

4 r^{2}

4 \times 14.0625 = 56.25

Rounding to the nearest integer, we arrive at the final result of 56.

Animated Solution for Mathematics - Circles: Let a point $P$ be such that its distance from the point $(5, 0)$ is thrice the distance of $P$ from the point $(- 5, 0)$ . If the locus of the point $P$ is a circle of radius $r$ , then $4 r^{2}$ is equal to (Round off to the nearest integer)

Enter Numerical Value:

Visualized Solution

$A (5, 0)$ and $B (- 5, 0)$

$P A = 3 P B$

$P A^{2} = 9 P B^{2}$

$d^{2} = (x_{2} - x_{1})^{2} + (y_{2} - y_{1})^{2}$

$(x - 5)^{2} + y^{2}$

$9 [(x + 5)^{2} + y^{2}]$

$8 x^{2} + 8 y^{2} + 100 x + 200 = 0$

$x^{2} + y^{2} + \frac{25}{2} x + 25 = 0$

$g, f, c$

$r^{2} = g^{2} + f^{2} - c$

$r^{2} = \frac{225}{16}$

$4 r^{2}$

$56.25 \approx 56$

The Sigma Insight: Standard and General Equation of a Circle

Analyzing the Setup

The Algebraic Foundation

The Simplification

Final Calculation

Similar Questions

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Enter Numerical Value:

Visualized Solution

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } A(5,0) and .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } B(−5,0)

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } PA=3PB

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } PA2=9PB2

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } d2=(x2​−x1​)2+(y2​−y1​)2

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } (x−5)2+y2

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } 9[(x+5)2+y2]

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } 8x2+8y2+100x+200=0

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } x2+y2+225​x+25=0

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } g,f,c

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } r2=g2+f2−c

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } r2=16225​

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } 4r2

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } 56.25≈56

The Sigma Insight: Standard and General Equation of a Circle

Analyzing the Setup

The Algebraic Foundation

The Simplification

Final Calculation

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$A (5, 0)$ and $B (- 5, 0)$

$P A = 3 P B$

$P A^{2} = 9 P B^{2}$

$d^{2} = (x_{2} - x_{1})^{2} + (y_{2} - y_{1})^{2}$

$(x - 5)^{2} + y^{2}$

$9 [(x + 5)^{2} + y^{2}]$

$8 x^{2} + 8 y^{2} + 100 x + 200 = 0$

$x^{2} + y^{2} + \frac{25}{2} x + 25 = 0$

$g, f, c$

$r^{2} = g^{2} + f^{2} - c$

$r^{2} = \frac{225}{16}$

$4 r^{2}$

$56.25 \approx 56$

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