JEE Main 2024 (06 Apr Shift 2)

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Animated Solution for Mathematics - Circles: If the locus of the point, whose distances from the point $(2, 1)$ and $(1, 3)$ are in the ratio $5 : 4$ , is $a x^{2} + b y^{2} + c x y + d x + ey + 170 = 0$ , then the value of $a^{2} + 2 b + 3 c + 4 d + e$ is equal to :

Select Answer:

Progressive Hints

Hint 1

Hint 1: Visualize the locus. You are looking for the set of all points

P (x, y)

such that the ratio of their distances from two fixed points

A (2, 1)

and

B (1, 3)

is a constant

5 : 4

. This is the classic definition of an Apollonian Circle. Do not try to find the center and radius directly; let the algebra guide you.

Hint 2

Hint 2: To avoid the complexity of square roots in the distance formula, square the ratio equation immediately. If

\frac{P A}{P B} = \frac{5}{4}

, then

16 \cdot P A^{2} = 25 \cdot P B^{2}

. This will transform the problem into a straightforward quadratic equation in

x

and

y

Hint 3

Hint 3: Expand the squared terms carefully. The equation becomes

16 [(x - 2)^{2} + (y - 1)^{2}] = 25 [(x - 1)^{2} + (y - 3)^{2}]

. Remember that

(x - a)^{2} = x^{2} - 2 a x + a^{2}

. Take your time to expand and group the

x^{2}

y^{2}

x

, and

y

terms to match the form

a x^{2} + b y^{2} + c x y + d x + ey + 170 = 0

Visualized Solution

Defining the Moving Point $P$

Let the moving point be $P (x, y)$ .
Given fixed points: $A (2, 1)$ and $B (1, 3)$ .
The locus is the set of all points $P$ satisfying the given distance ratio.

Setting up the Distance Ratio

Condition: $\frac{P A}{P B} = \frac{5}{4}$
This implies $4 \cdot P A = 5 \cdot P B$ .

Squaring to Remove Square Roots

To simplify, square both sides:
$(\frac{P A}{P B})^{2} = (\frac{5}{4})^{2}$
$\frac{P A ^{2}}{P B ^{2}} = \frac{25}{16}$
$16 \cdot P A^{2} = 25 \cdot P B^{2}$

Applying the Distance Formula

Using Distance Formula: $d^{2} = (x_{2} - x_{1})^{2} + (y_{2} - y_{1})^{2}$
$P A^{2} = (x - 2)^{2} + (y - 1)^{2}$
$P B^{2} = (x - 1)^{2} + (y - 3)^{2}$

Substituting into the Equation

Substitute into the squared ratio equation:
$16 [(x - 2)^{2} + (y - 1)^{2}] = 25 [(x - 1)^{2} + (y - 3)^{2}]$

Expanding the Quadratic Terms

Expand $(x - 2)^{2}$ and $(y - 1)^{2}$ etc.:
$16 (x^{2} - 4 x + 4 + y^{2} - 2 y + 1) = 25 (x^{2} - 2 x + 1 + y^{2} - 6 y + 9)$

Simplifying Inside the Brackets

Combine constants inside brackets:
$16 (x^{2} + y^{2} - 4 x - 2 y + 5) = 25 (x^{2} + y^{2} - 2 x - 6 y + 10)$

Distributing the Coefficients

Multiply across:
$16 x^{2} + 16 y^{2} - 64 x - 32 y + 80 = 25 x^{2} + 25 y^{2} - 50 x - 150 y + 250$

Grouping and Rearranging Terms

Rearrange to form $A x^{2} + B y^{2} + C x y + D x + E y + F = 0$ :
$(25 - 16) x^{2} + (25 - 16) y^{2} + (- 50 + 64) x + (- 150 + 32) y + (250 - 80) = 0$
$9 x^{2} + 9 y^{2} + 14 x - 118 y + 170 = 0$

Identifying the Coefficients

Compare $9 x^{2} + 9 y^{2} + 14 x - 118 y + 170 = 0$ with $a x^{2} + b y^{2} + c x y + d x + ey + 170 = 0$ :
$a = 9$
$b = 9$
$c = 0$
$d = 14$
$e = - 118$

Setting up the Final Expression

Calculate: $a^{2} + 2 b + 3 c + 4 d + e$
$= 9^{2} + 2 (9) + 3 (0) + 4 (14) + (- 118)$

Computing the Final Value

$= 81 + 18 + 0 + 56 - 118$
$= 155 - 118 = 37$
Final Answer: $37$

00:00 / 00:00

The Sigma Insight: Standard and General Equation of a Circle

Solution Diagram

Analyzing the Setup

Welcome to a beautiful journey through coordinate geometry. Today, we are tracing the path of a point

P (x, y)

that is constrained by a specific geometric rule.

We have two fixed anchors,

A (2, 1)

and

B (1, 3)

. The point

P

moves such that its distance from

A

and its distance from

B

maintain a constant ratio of

5 : 4

This is a classic locus problem, specifically defining an Apollonian Circle. We seek the equation of the path that

P

carves out in the plane.

The Algebraic Strategy

The given condition is

\frac{P A}{P B} = \frac{5}{4}

. Using the standard distance formula, we express this as:

\frac{( x - 2 ) ^{2} + ( y - 1 ) ^{2}}{( x - 1 ) ^{2} + ( y - 3 ) ^{2}} = \frac{5}{4}

To eliminate the radicals, we square both sides of the equation. This yields the relation

16 \cdot P A^{2} = 25 \cdot P B^{2}

Substituting the distance formula into this squared relation, we obtain:

16 [(x - 2)^{2} + (y - 1)^{2}] = 25 [(x - 1)^{2} + (y - 3)^{2}]

The Expansion

A Test of Precision

We now expand the squares on both sides. On the left, we have:

16 (x^{2} - 4 x + 4 + y^{2} - 2 y + 1) = 16 (x^{2} + y^{2} - 4 x - 2 y + 5)

On the right, we have:

25 (x^{2} - 2 x + 1 + y^{2} - 6 y + 9) = 25 (x^{2} + y^{2} - 2 x - 6 y + 10)

Distributing the coefficients, we arrive at:

16 x^{2} + 16 y^{2} - 64 x - 32 y + 80 = 25 x^{2} + 25 y^{2} - 50 x - 150 y + 250

By moving all terms to one side to group like terms, we get:

(25 - 16) x^{2} + (25 - 16) y^{2} + (- 50 + 64) x + (- 150 + 32) y + (250 - 80) = 0

This simplifies to the elegant equation:

9 x^{2} + 9 y^{2} + 14 x - 118 y + 170 = 0

The Final Victory

We compare our result with the given form

a x^{2} + b y^{2} + c x y + d x + ey + 170 = 0

. We identify the coefficients as

a = 9

b = 9

c = 0

d = 14

, and

e = - 118

The final task is to compute the value of

a^{2} + 2 b + 3 c + 4 d + e

. Substituting our identified values:

9^{2} + 2 (9) + 3 (0) + 4 (14) + (- 118)

Calculating each term, we get

81 + 18 + 0 + 56 - 118

. Summing the positive values results in

155

Subtracting

118

from

155

, we arrive at the final result:

Common Traps & FAQs

Wait, why do we square the ratio? Can't we just use the distance formula with square roots?

You absolutely could, but that is a path filled with unnecessary pain! Using square roots would lead to radical equations that are difficult to solve and prone to algebraic errors. By squaring both sides, we eliminate the radicals and convert the problem into a standard polynomial equation, which is much cleaner and safer to manipulate.

Why is the locus a circle? Is there a shortcut?

Yes, there is! The locus of a point whose distances from two fixed points are in a constant ratio

k \neq = 1

is always a circle, known as the Apollonian Circle. While you could use this property to find the center and radius, for this specific JEE problem, the algebraic method is more direct and less prone to conceptual errors.

I got

9 x^{2} + 9 y^{2} + 14 x - 118 y + 170 = 0

. Did I miss the

x y

term?

You did not miss it! In this specific case, the coefficient of the

x y

term is indeed zero. This is perfectly normal for a circle whose axes are parallel to the coordinate axes. Your result is correct, and it confirms that the locus is a circle.

Animated Solution for Mathematics - Circles: If the locus of the point, whose distances from the point $(2, 1)$ and $(1, 3)$ are in the ratio $5 : 4$ , is $a x^{2} + b y^{2} + c x y + d x + ey + 170 = 0$ , then the value of $a^{2} + 2 b + 3 c + 4 d + e$ is equal to :

Select Answer:

Visualized Solution

Defining the Moving Point $P$

Setting up the Distance Ratio

Squaring to Remove Square Roots

Applying the Distance Formula

Substituting into the Equation

Expanding the Quadratic Terms

Simplifying Inside the Brackets

Distributing the Coefficients

Grouping and Rearranging Terms

Identifying the Coefficients

Setting up the Final Expression

Computing the Final Value

The Sigma Insight: Standard and General Equation of a Circle

Analyzing the Setup

The Algebraic Strategy

The Expansion

The Final Victory

Similar Questions

Let $A$ be the point $(1, 2)$ and $B$ be any point on the curve $x^{2} + y^{2} = 16$ . If the centre of the locus of the point $P$ , which divides the line segment $A B$ in the ratio $3 : 2$ is the point $C (α, β)$ , then the length of the line segment $A C$ is

A line segment $A B$ of length $λ$ moves such that the points $A$ and $B$ remain on the periphery of a circle of radius $λ$ . Then the locus of the point, that divides the line segment $A B$ in the ratio $2 : 3$ , is a circle of radius

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)

The locus of a point, which moves such that the sum of squares of its distances from the points $(0, 0), (1, 0), (0, 1), (1, 1)$ is 18 units, is a circle of diameter $d$ . Then $d^{2}$ is equal to .

If the locus of the mid-point of the line segment from the point $(3, 2)$ to a point on the circle, $x^{2} + y^{2} = 1$ is a circle of the radius $r$ , then $r$ is equal to :

A circle passes through the points (2,3) and (4,5). If its centre lies on the line, $y - 4 x + 3 = 0$ , then its radius is equal to

A circle is given by $x^{2} + (y - 1)^{2} = 1$ , another circle $C$ touches it externally and also the x-axis, then the locus of its centre is

Let a circle passing through $(2, 0)$ have its centre at the point $(h, k)$ . Let $(x_{c}, y_{c})$ be the point of intersection of the lines $3 x + 5 y = 1$ and $(2 + c) x + 5 c^{2} y = 1$ . If $h = lim_{c \to 1} x_{c}$ and $k = lim_{c \to 1} y_{c}$ , then the equation of the circle is :

Select Answer:

Visualized Solution

Defining the Moving Point .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } P

Setting up the Distance Ratio

Squaring to Remove Square Roots

Applying the Distance Formula

Substituting into the Equation

Expanding the Quadratic Terms

Simplifying Inside the Brackets

Distributing the Coefficients

Grouping and Rearranging Terms

Identifying the Coefficients

Setting up the Final Expression

Computing the Final Value

The Sigma Insight: Standard and General Equation of a Circle

Analyzing the Setup

The Algebraic Strategy

The Expansion

The Final Victory

Similar Questions

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If the locus of the mid-point of the line segment from the point (3,2) to a point on the circle, x2+y2=1 is a circle of the radius r, then r is equal to :

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } A circle passes through the points (2,3) and (4,5). If its centre lies on the line, y−4x+3=0, then its radius is equal to

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } A circle is given by x2+(y−1)2=1, another circle C touches it externally and also the x-axis, then the locus of its centre is

Defining the Moving Point $P$

If the locus of the mid-point of the line segment from the point $(3, 2)$ to a point on the circle, $x^{2} + y^{2} = 1$ is a circle of the radius $r$ , then $r$ is equal to :

A circle passes through the points (2,3) and (4,5). If its centre lies on the line, $y - 4 x + 3 = 0$ , then its radius is equal to

A circle is given by $x^{2} + (y - 1)^{2} = 1$ , another circle $C$ touches it externally and also the x-axis, then the locus of its centre is