JEE Advanced 1993

LEVELJEE Main

Animated Solution for Mathematics - Circles: 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 .........

Q: Do I really need to divide the original equation by 4 at the very beginning?

Absolutely! The standard form of a circle is $x^2 + y^2 + 2gx + 2fy + c = 0$. If the coefficients of $x^2$ and $y^2$ are not 1, you cannot directly identify the center $(-g, -f)$ or the radius $\sqrt{g^2 + f^2 - c}$. Cleaning up the equation is not just a suggestion; it is a vital step to avoid errors.

Q: Why is the angle halved in the triangle $ACM$?

Great observation! The triangle formed by the center $C$ and the chord $AB$ is an isosceles triangle because $CA = CB = r$. When we draw the perpendicular from the center $C$ to the mid-point $M$ of the chord $AB$, this line segment $CM$ acts as the angle bisector of $\angle ACB$. Therefore, $\angle ACM$ is exactly half of $\angle ACB$.

Progressive Hints

Hint 1

Hint 1: Visualize the geometry. The locus of the mid-points of chords that subtend a constant angle at the center of a circle is itself a circle. Why? Because the distance from the center to the mid-point of such a chord is constant. Can you see why this distance is fixed?

Hint 2

Hint 2: Use the right-angled triangle formed by the center of the circle, the mid-point of the chord, and one of the endpoints of the chord. If the chord subtends an angle of

2 π /3

at the center, what is the angle in this right-angled triangle? Use trigonometry to find the distance from the center to the mid-point.

Hint 3

Hint 3: Once you have the constant distance

d

from the center

C (h_{c}, k_{c})

to the mid-point

M (x, y)

, the equation of the locus is simply

(x - h_{c})^{2} + (y - k_{c})^{2} = d^{2}

. Substitute the center coordinates and the value of

d

to find the final equation.

Visualized Solution

Standardizing the Circle Equation

Given equation: $4 x^{2} + 4 y^{2} - 12 x + 4 y + 1 = 0$
Divide by $4$ to get the standard form:
$x^{2} + y^{2} - 3 x + y + \frac{1}{4} = 0$

Finding Center $C$ and Radius $r$

Comparing with $x^{2} + y^{2} + 2 g x + 2 f y + c = 0$ :
$2 g = - 3 ⟹ g = - \frac{3}{2}$
$2 f = 1 ⟹ f = \frac{1}{2}$
Center $C = (- g, - f) = (\frac{3}{2}, - \frac{1}{2})$
Radius $r = g^{2} + f^{2} - c = \frac{9}{4} + \frac{1}{4} - \frac{1}{4} = \frac{3}{2}$

Geometric Setup of the Chord

Let $M (h, k)$ be the mid-point of chord $A B$ .
Given $∠ A C B = \frac{2 π}{3} = 12 0^{\circ}$ .

Analyzing $△ A C M$

Join $C$ to $M$ . Since $M$ is the mid-point, $C M ⊥ A B$ .
$△ A C M$ is a right-angled triangle.
$∠ A C M = \frac{1}{2} ∠ A C B = 6 0^{\circ}$ .

Trigonometric Relation in $△ A C M$

In right $△ A C M$ :
$cos (6 0^{\circ}) = \frac{Base}{Hypotenuse} = \frac{C M}{A C}$
Substitute knowns: $\frac{1}{2} = \frac{C M}{3/2}$

Calculating the Distance $C M$

$C M = \frac{3}{2} \times \frac{1}{2}$
$C M = \frac{3}{4}$

Applying the Distance Formula

Distance between $C (\frac{3}{2}, - \frac{1}{2})$ and $M (h, k)$ is $\frac{3}{4}$ .
$C M = (h - \frac{3}{2})^{2} + (k - (- \frac{1}{2}))^{2} = \frac{3}{4}$

Squaring the Equation

Square both sides to remove the square root:
$(h - \frac{3}{2})^{2} + (k + \frac{1}{2})^{2} = (\frac{3}{4})^{2}$
$(h - \frac{3}{2})^{2} + (k + \frac{1}{2})^{2} = \frac{9}{16}$

Algebraic Expansion

Expand the squares:
$(h^{2} - 3 h + \frac{9}{4}) + (k^{2} + k + \frac{1}{4}) = \frac{9}{16}$
Combine constant terms on the left:
$h^{2} + k^{2} - 3 h + k + \frac{10}{4} = \frac{9}{16}$

Simplifying the Equation

Multiply the entire equation by $16$ to clear denominators:
$16 (h^{2} + k^{2} - 3 h + k + \frac{5}{2}) = 16 (\frac{9}{16})$
$16 h^{2} + 16 k^{2} - 48 h + 16 k + 40 = 9$
$16 h^{2} + 16 k^{2} - 48 h + 16 k + 31 = 0$

Final Locus Equation

To find the locus, replace $(h, k)$ with $(x, y)$ :
$16 x^{2} + 16 y^{2} - 48 x + 16 y + 31 = 0$
Conclusion: The locus of the mid-points is a concentric circle with the same center but a smaller radius.

00:00 / 00:00

The Sigma Insight: Standard and General Equation of a Circle

Solution Diagram

Analyzing the Setup

We begin with the given circle equation:

4 x^{2} + 4 y^{2} - 12 x + 4 y + 1 = 0

To simplify, we divide the entire equation by

4

x^{2} + y^{2} - 3 x + y + \frac{1}{4} = 0

By comparing this to the general form

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

, we identify the center

C

and the radius

r

. The center is

C (\frac{3}{2}, - \frac{1}{2})

and the radius is

r = \frac{3}{2}

The Hidden Triangle

Consider a chord

A B

in this circle with mid-point

M (h, k)

. The chord subtends an angle of

\frac{2 π}{3}

(

12 0^{\circ}

) at the center

C

Connecting the center

C

to the mid-point

M

creates a right-angled triangle

△ A C M

. Since

C M

is the perpendicular bisector of chord

A B

, it bisects the central angle. Thus,

∠ A C M = 6 0^{\circ}

In the right-angled triangle

△ A C M

, we apply trigonometry:

cos (6 0^{\circ}) = \frac{C M}{A C}

Given

cos (6 0^{\circ}) = \frac{1}{2}

and

A C = r = \frac{3}{2}

, we calculate the distance

C M

C M = \frac{3}{2} \times \frac{1}{2} = \frac{3}{4}

The Final Algebraic Synthesis

The mid-point

M (h, k)

maintains a constant distance of

\frac{3}{4}

from the center

C (\frac{3}{2}, - \frac{1}{2})

. Using the distance formula, we establish the locus:

(h - \frac{3}{2})^{2} + (k + \frac{1}{2})^{2} = \frac{3}{4}

Squaring both sides yields:

(h - \frac{3}{2})^{2} + (k + \frac{1}{2})^{2} = \frac{9}{16}

Expanding the terms, we obtain:

h^{2} - 3 h + \frac{9}{4} + k^{2} + k + \frac{1}{4} = \frac{9}{16}

Combining the constants

\frac{9}{4} + \frac{1}{4} = \frac{10}{4} = \frac{5}{2}

, the equation becomes:

h^{2} + k^{2} - 3 h + k + \frac{5}{2} = \frac{9}{16}

Multiplying the entire equation by

16

to clear the denominators:

16 h^{2} + 16 k^{2} - 48 h + 16 k + 40 = 9

Subtracting

9

from both sides, we arrive at the final locus equation: $16 x^{2} + 16 y^{2} - 48 x + 16 y + 31 = 0$

Common Traps & FAQs

Wait, why is the locus a circle? Can't it be a line or something else?

That is a fantastic question! Think about the definition of a circle: it is the set of all points at a fixed distance from a center. In our problem, the distance from the center

C

to the mid-point

M

is fixed by the angle the chord subtends. Since the distance

C M

is constant, the mid-point

M

must trace out a circle centered at

C

with radius

C M

Do I really need to divide the original equation by 4 at the very beginning?

Absolutely! The standard form of a circle is

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

. If the coefficients of

x^{2}

and

y^{2}

are not 1, you cannot directly identify the center

(- g, - f)

or the radius

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

. Cleaning up the equation is not just a suggestion; it is a vital step to avoid errors.

Why is the angle halved in the triangle

A C M

Great observation! The triangle formed by the center

C

and the chord

A B

is an isosceles triangle because

C A = C B = r

. When we draw the perpendicular from the center

C

to the mid-point

M

of the chord

A B

, this line segment

C M

acts as the angle bisector of

∠ A C B

. Therefore,

∠ A C M

is exactly half of

∠ A C B

Animated Solution for Mathematics - Circles: 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 .........

Visualized Solution

Standardizing the Circle Equation

Finding Center $C$ and Radius $r$

Geometric Setup of the Chord

Analyzing $△ A C M$

Trigonometric Relation in $△ A C M$

Calculating the Distance $C M$

Applying the Distance Formula

Squaring the Equation

Algebraic Expansion

Simplifying the Equation

Final Locus Equation

The Sigma Insight: Standard and General Equation of a Circle

Analyzing the Setup

The Hidden Triangle

The Final Algebraic Synthesis

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 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

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 .........

Let a circle $C$ touch the lines $L_{1} : 4 x - 3 y + K_{1} = 0$ and $L_{2} : 4 x - 3 y + K_{2} = 0, K_{1}, K_{2} \in R$ . If a line passing through the centre of the circle $C$ intersects $L_{1}$ at $(- 1, 2)$ and $L_{2}$ at $(3, - 6)$ , then the equation of the circle $C$ is

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 :

Let the locus of the mid points of the chords of circle $x^{2} + (y - 1)^{2} = 1$ drawn from the origin intersect the line $x + y = 1$ at $P$ and $Q$ . Then, the length of $P Q$ is :

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

Through a fixed point $(h, k)$ secants are drawn to the circle $x^{2} + y^{2} = r^{2}$ . Show that the locus of the mid-points of the secants intercepted by the circle is $x^{2} + y^{2} = h x + k y$ .

Let a circle of radius 4 pass through the origin $O$ , the points $A (- 3 a, 0)$ and $B (0, - 2 b)$ , where $a$ and $b$ are real parameters and $ab \neq = 0$ . Then the locus of the centroid of $△ O A B$ is a circle of radius

Visualized Solution

Standardizing the Circle Equation

Finding Center .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } C and Radius .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } r

Geometric Setup of the Chord

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

Trigonometric Relation in .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } △ACM

Calculating the Distance .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } CM

Applying the Distance Formula

Squaring the Equation

Algebraic Expansion

Simplifying the Equation

Final Locus Equation

The Sigma Insight: Standard and General Equation of a Circle

Analyzing the Setup

The Hidden Triangle

The Final Algebraic Synthesis

Similar Questions

.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; } From the point A(0,3) on the circle x2+4x+(y−3)2=0, a chord AB is drawn and extended to a point M such that AM=2AB. The equation of the locus of M is .........

.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; } Let the locus of the mid points of the chords of circle x2+(y−1)2=1 drawn from the origin intersect the line x+y=1 at P and Q. Then, the length of PQ is :

.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

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Through a fixed point (h,k) secants are drawn to the circle x2+y2=r2. Show that the locus of the mid-points of the secants intercepted by the circle is x2+y2=hx+ky.

Finding Center $C$ and Radius $r$

Analyzing $△ A C M$

Trigonometric Relation in $△ A C M$

Calculating the Distance $C M$

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

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 .........

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 :

Let the locus of the mid points of the chords of circle $x^{2} + (y - 1)^{2} = 1$ drawn from the origin intersect the line $x + y = 1$ at $P$ and $Q$ . Then, the length of $P Q$ is :

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

Through a fixed point $(h, k)$ secants are drawn to the circle $x^{2} + y^{2} = r^{2}$ . Show that the locus of the mid-points of the secants intercepted by the circle is $x^{2} + y^{2} = h x + k y$ .