JEE Advanced 1986

LEVELJEE Main

Animated Solution for Mathematics - Circles: From the point $A (0, 3)$ on the circle $x^{2} + 4 x + (y - 3)^{2} = 0$ , a chord $A B$ is drawn and extended to a point $M$ such that $A M = 2 A B$ . The equation of the locus of $M$ is .........

Q: Wait, why is $B$ the midpoint of $AM$? I thought it was just on the line.

Great question! The problem states $AM = 2AB$. Since $M$ is an extension of the chord $AB$, the points $A, B,$ and $M$ are collinear in that order. So, $AM = AB + BM$. Substituting this into the given equation, we get $AB + BM = 2AB$, which simplifies to $BM = AB$. Since $B$ is between $A$ and $M$ and the distances $AB$ and $BM$ are equal, $B$ must be the midpoint.

Q: Why do we replace $(h, k)$ with $(x, y)$ at the end?

That is a standard convention in locus problems. We use $(h, k)$ as a placeholder for an arbitrary point $M$ that satisfies the geometric condition. Once we have the relationship between $h$ and $k$, we replace them with $(x, y)$ to represent the general equation of the curve (the locus) that all such points $M$ must lie on.

Visualized Solution

Analyzing the Given Circle Equation

Given circle equation: $x^{2} + 4 x + (y - 3)^{2} = 0$
Let's rewrite it in standard form: $(x + 2)^{2} - 4 + (y - 3)^{2} = 0 ⟹ (x + 2)^{2} + (y - 3)^{2} = 2^{2}$
This represents a circle with center $C_{1} (- 2, 3)$ and radius $R = 2$ .
The given point is $A (0, 3)$ . Let's verify if $A$ lies on the circle: $(0 + 2)^{2} + (3 - 3)^{2} = 4$ , which is true!

Defining Chord $A B$ and Extension to $M$

A chord $A B$ is drawn from point $A (0, 3)$ to another point $B$ on the circle.
This chord is extended to a point $M (h, k)$ such that $A M = 2 A B$ .
This geometric condition means that $B$ is exactly halfway between $A$ and $M$ .

Applying the Midpoint Formula

Since $A M = 2 A B$ , point $B$ is the midpoint of segment $A M$ .
Let $A = (0, 3)$ and $M = (h, k)$ .
Using the midpoint formula, the coordinates of $B$ are:
$B = \left(\frac{0 + h}{2}, \frac{3 + k}{2}\right) = \left(\frac{h}{2}, \frac{k + 3}{2} ight)$

Point $B$ Lies on the Circle

Point $B\left(\frac{h}{2}, \frac{k + 3}{2} ight)$ must satisfy the equation of the given circle.
Circle equation: $x^{2} + 4 x + (y - 3)^{2} = 0$
We will substitute $x = \frac{h}{2}$ and $y = \frac{k + 3}{2}$ into this equation.

Substituting Coordinates of $B$

Substituting $x = \frac{h}{2}$ and $y = \frac{k + 3}{2}$ :
$\left(\frac{h}{2}\right)^2 + 4\left(\frac{h}{2} ight) + \left(\frac{k + 3}{2} - 3\right)^2 = 0$
Let's focus on simplifying the terms one by one.

Simplifying the Terms

First term: $(\frac{h}{2})^{2} = \frac{h ^{2}}{4}$
Second term: $4\left(\frac{h}{2} ight) = 2h$
Third term: $\left(\frac{k + 3}{2} - 3\right)^2 = \left(\frac{k + 3 - 6}{2} ight)^2 = \left(\frac{k - 3}{2} ight)^2$

Reassembling the Equation

Substitute the simplified terms back:
$\frac{h ^{2}}{4} + 2 h + \frac{( k - 3 ) ^{2}}{4} = 0$
Expand the numerator of the third term: $(k - 3)^{2} = k^{2} - 6 k + 9$
The equation becomes: $\frac{h ^{2}}{4} + 2 h + \frac{k ^{2} - 6 k + 9}{4} = 0$

Clearing the Fractions

To eliminate the denominators, multiply the entire equation by $4$ :
$4 \times (\frac{h ^{2}}{4} + 2 h + \frac{k ^{2} - 6 k + 9}{4}) = 4 \times 0$
$h^{2} + 8 h + k^{2} - 6 k + 9 = 0$
Rearranging terms: $h^{2} + k^{2} + 8 h - 6 k + 9 = 0$

Replacing $(h, k)$ with $(x, y)$

To find the general locus of point $M$ , replace $h$ with $x$ and $k$ with $y$ :
Locus Equation: $x^{2} + y^{2} + 8 x - 6 y + 9 = 0$
This is a circle with center $(- 4, 3)$ and radius $4$ .

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The Sigma Insight: Standard and General Equation of a Circle

Solution Diagram

Analyzing the Setup

The motion occurs on a coordinate plane where point

B

moves along a circle defined by the equation

x^{2} + 4 x + (y - 3)^{2} = 0

By completing the square for the

x

terms, we rewrite the equation as:

(x + 2)^{2} - 4 + (y - 3)^{2} = 0

This simplifies to the standard form:

(x + 2)^{2} + (y - 3)^{2} = 2^{2}

We are observing a circle centered at

C_{1} (- 2, 3)

with a radius of

2

. Our fixed anchor point is

A (0, 3)

, which lies on the circumference of this circle.

The Geometric Transformation

A chord

A B

is drawn from

A

to a point

B

on the circle and extended to a point

M (h, k)

such that

A M = 2 A B

Because the total length

A M

is twice the length of

A B

, the segment

B M

must be equal to

A B

. Since

A

B

, and

M

are collinear,

B

is the midpoint of the segment

A M

Using the midpoint formula, we express the coordinates of

B

in terms of

M (h, k)

and

A (0, 3)

B = (\frac{0 + h}{2}, \frac{3 + k}{2}) = (\frac{h}{2}, \frac{k + 3}{2})

The Master Equation

Since point

B

must lie on the original circle, its coordinates must satisfy the equation

x^{2} + 4 x + (y - 3)^{2} = 0

. Substituting

x = \frac{h}{2}

and

y = \frac{k + 3}{2}

into this equation, we obtain:

(\frac{h}{2})^{2} + 4 (\frac{h}{2}) + (\frac{k + 3}{2} - 3)^{2} = 0

Simplifying the terms step-by-step: 1. The first term is

\frac{h ^{2}}{4}

. 2. The second term,

4 (\frac{h}{2})

, simplifies to

2 h

. 3. The third term becomes

(\frac{k + 3 - 6}{2})^{2} = \frac{( k - 3 ) ^{2}}{4}

Combining these, we have:

\frac{h ^{2}}{4} + 2 h + \frac{( k - 3 ) ^{2}}{4} = 0

Final Calculation

To clear the fractions, we multiply the entire equation by

4

h^{2} + 8 h + (k - 3)^{2} = 0

Expanding the squared term

(k - 3)^{2} = k^{2} - 6 k + 9

, the equation becomes:

h^{2} + k^{2} + 8 h - 6 k + 9 = 0

Replacing

(h, k)

with

(x, y)

, we find the locus of $M$ : $x^{2} + y^{2} + 8 x - 6 y + 9 = 0$

This result represents a circle centered at

(- 4, 3)

with a radius of

4

. We have successfully mapped the path of

M

B

moves along the original circle.

Common Traps & FAQs

Wait, why is

B

the midpoint of

A M

? I thought it was just on the line.

Great question! The problem states

A M = 2 A B

. Since

M

is an extension of the chord

A B

, the points

A, B,

and

M

are collinear in that order. So,

A M = A B + B M

. Substituting this into the given equation, we get

A B + B M = 2 A B

, which simplifies to

B M = A B

. Since

B

is between

A

and

M

and the distances

A B

and

B M

are equal,

B

must be the midpoint.

Can I use polar coordinates to solve this?

You absolutely could, but it might be overkill! While polar coordinates are powerful for circle problems, the Cartesian approach using the midpoint formula is much more direct here. Stick to the simplest path unless the geometry suggests otherwise.

Why do we replace

(h, k)

with

(x, y)

at the end?

That is a standard convention in locus problems. We use

(h, k)

as a placeholder for an arbitrary point

M

that satisfies the geometric condition. Once we have the relationship between

h

and

k

, we replace them with

(x, y)

to represent the general equation of the curve (the locus) that all such points

M

must lie on.

Animated Solution for Mathematics - Circles: From the point $A (0, 3)$ on the circle $x^{2} + 4 x + (y - 3)^{2} = 0$ , a chord $A B$ is drawn and extended to a point $M$ such that $A M = 2 A B$ . The equation of the locus of $M$ is .........

Visualized Solution

Analyzing the Given Circle Equation

Defining Chord $A B$ and Extension to $M$

Applying the Midpoint Formula

Point $B$ Lies on the Circle

Substituting Coordinates of $B$

Simplifying the Terms

Reassembling the Equation

Clearing the Fractions

Replacing $(h, k)$ with $(x, y)$

The Sigma Insight: Standard and General Equation of a Circle

Analyzing the Setup

The Geometric Transformation

The Master Equation

Final Calculation

Similar Questions

Let $C$ be the circle with centre $(0, 0)$ and radius 3 units. The equation of the locus of the mid points of the chords of the circle $C$ that subtend an angle of $\frac{2 π}{3}$ at its center is

The equation of the locus of the mid-points of the circle $4 x^{2} + 4 y^{2} - 12 x + 4 y + 1 = 0$ that subtend an angle of $2 π /3$ at its centre is .........

The locus of the mid-point of a chord of the circle $x^{2} + y^{2} = 4$ which subtends a right angle at the origin is

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The locus of the centre of a circle, which touches externally the circle $x^{2} + y^{2} - 6 x - 8 y + 14 = 0$ and also touches the y-axis, is given by the equation:

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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

Visualized Solution

Analyzing the Given Circle Equation

Defining Chord .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } AB and Extension to .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } M

Applying the Midpoint Formula

Point .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } B Lies on the Circle

Substituting Coordinates of .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } B

Simplifying the Terms

Reassembling the Equation

Clearing the Fractions

Replacing .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } (h,k) with .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } (x,y)

The Sigma Insight: Standard and General Equation of a Circle

Analyzing the Setup

The Geometric Transformation

The Master Equation

Final Calculation

Similar Questions

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.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The locus of the mid-point of a chord of the circle x2+y2=4 which subtends a right angle at the origin is

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } A variable circle passes through the fixed point A(p,q) and touches x-axis. The locus of the other end of the diameter through A is

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } A circle cuts a chord of length 4a on the x-axis and passes through a point on the y-axis, distant 2b from the origin. Then the locus of the centre of this circle, is:

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The locus of the centre of a circle, which touches externally the circle x2+y2−6x−8y+14=0 and also touches the y-axis, is given by the equation:

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Defining Chord $A B$ and Extension to $M$

Point $B$ Lies on the Circle

Substituting Coordinates of $B$

Replacing $(h, k)$ with $(x, y)$

The equation of the locus of the mid-points of the circle $4 x^{2} + 4 y^{2} - 12 x + 4 y + 1 = 0$ that subtend an angle of $2 π /3$ at its centre is .........

The locus of the mid-point of a chord of the circle $x^{2} + y^{2} = 4$ which subtends a right angle at the origin is

A variable circle passes through the fixed point $A (p, q)$ and touches x-axis. The locus of the other end of the diameter through $A$ is

A circle cuts a chord of length $4 a$ on the x-axis and passes through a point on the y-axis, distant $2 b$ from the origin. Then the locus of the centre of this circle, is:

The locus of the centre of a circle, which touches externally the circle $x^{2} + y^{2} - 6 x - 8 y + 14 = 0$ and also touches the y-axis, is given by the equation:

Find the equations of the circle passing through $(- 4, 3)$ and touching the lines $x + y = 2$ and $x - y = 2$ .

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