JEE Advanced 1998

LEVELJEE Main

Animated Solution for Mathematics - Circles: $C_{1}$ and $C_{2}$ are two concentric circles, the radius of $C_{2}$ being twice that of $C_{1}$ . From a point $P$ on $C_{2}$ , tangents $P A$ and $P B$ are drawn to $C_{1}$ . Prove that the centroid of the triangle $P A B$ lies on $C_{1}$ .

Q: I always forget the Chord of Contact formula. Is there another way?

You could use the slope-point form of a line $y - y_1 = m(x - x_1)$ and the condition of tangency $c^2 = r^2(1+m^2)$, but that is a long, winding road. In JEE Advanced, time is currency. Memorizing $T=0$ is a high-yield investment. It works for any conic section, not just circles! It is the single most efficient tool for problems involving tangents from an external point.

Progressive Hints

Hint 1

Don't get lost in general variables. The problem possesses rotational symmetry. What happens if you align the center of the circles with the origin

(0, 0)

and place point

P

on the x-axis? Does the geometry change if you rotate the entire system?

Hint 2

You are dealing with tangents from an external point to a circle. Do not derive the equations of the lines from scratch using slopes. Use the powerful 'Chord of Contact' formula,

T = 0

, which for the circle

x^{2} + y^{2} = r^{2}

and point

(x_{1}, y_{1})

x x_{1} + y y_{1} = r^{2}

Hint 3

Once you have the equation of the chord

A B

, find the intersection points

A

and

B

by solving the system of equations formed by the chord and the inner circle. Finally, apply the centroid formula

G = (\frac{x _{P} + x _{A} + x _{B}}{3}, \frac{y _{P} + y _{A} + y _{B}}{3})

and check if the result satisfies the equation of

C_{1}

Visualized Solution

Setting up the Circles $C_{1}$ and $C_{2}$

Let the center of both circles be the origin $(0, 0)$ .
Equation of $C_{1}$ : $x^{2} + y^{2} = r^{2}$
Equation of $C_{2}$ : $x^{2} + y^{2} = (2 r)^{2} = 4 r^{2}$

Defining Point $P$ on $C_{2}$

Let point $P$ lie on the x-axis for simplicity.
Since $P$ is on $C_{2}$ , its coordinates are $(2 r, 0)$ .

Drawing Tangents $P A$ and $P B$

Tangents $P A$ and $P B$ are drawn from $P$ to the inner circle $C_{1}$ .

The Chord of Contact $A B$

The line segment joining the points of tangency, $A$ and $B$ , is called the Chord of Contact.

Equation of the Chord of Contact

The equation of the chord of contact from an external point $(x_{1}, y_{1})$ to $x^{2} + y^{2} = r^{2}$ is $T = 0$ .
This gives $x x_{1} + y y_{1} = r^{2}$ .

Substituting Point $P$ into the Equation

Substitute $P (2 r, 0)$ into $x x_{1} + y y_{1} = r^{2}$ .
$x (2 r) + y (0) = r^{2}$

Simplifying the Chord Equation

$2 r x = r^{2} ⟹ x = \frac{r}{2}$
This means the chord of contact is a vertical line.

Finding the y-coordinates of $A$ and $B$

Points $A$ and $B$ lie on $C_{1}$ ( $x^{2} + y^{2} = r^{2}$ ) and the line $x = \frac{r}{2}$ .
Substitute $x = \frac{r}{2}$ into the circle's equation:
$(\frac{r}{2})^{2} + y^{2} = r^{2}$

Solving for $y$

$\frac{r ^{2}}{4} + y^{2} = r^{2} ⟹ y^{2} = r^{2} - \frac{r ^{2}}{4} = \frac{3 r ^{2}}{4}$
Therefore, $y = \pm \frac{r 3}{2}$

Coordinates of $A$ and $B$

Point $A = (\frac{r}{2}, \frac{r 3}{2})$
Point $B = (\frac{r}{2}, - \frac{r 3}{2})$

Formula for the Centroid $G$

The centroid $G (x_{G}, y_{G})$ of $△ P A B$ is given by:
$x_{G} = \frac{x _{P} + x _{A} + x _{B}}{3}$
$y_{G} = \frac{y _{P} + y _{A} + y _{B}}{3}$

Calculating the x-coordinate of $G$

$x_{G} = \frac{2 r + \frac{r}{2} + \frac{r}{2}}{3}$
$x_{G} = \frac{2 r + r}{3} = \frac{3 r}{3} = r$

Calculating the y-coordinate of $G$

$y_{G} = \frac{0 + \frac{r 3}{2} - \frac{r 3}{2}}{3}$
$y_{G} = \frac{0}{3} = 0$
So, $G = (r, 0)$

Final Verification

Check if $G (r, 0)$ lies on $C_{1}$ ( $x^{2} + y^{2} = r^{2}$ ).
Substitute $x = r, y = 0$ : $r^{2} + 0^{2} = r^{2}$
LHS = RHS. Hence Proved.

00:00 / 00:00

The Sigma Insight: Equation of Tangent and Normal

Solution Diagram

Analyzing the Setup

In coordinate geometry, the choice of your origin is your most powerful decision. Let us place the common center of our circles at the origin

(0, 0)

The inner circle

C_{1}

has a radius

r

, so its equation is

x^{2} + y^{2} = r^{2}

. The outer circle

C_{2}

has a radius of

2 r

, making its equation

x^{2} + y^{2} = 4 r^{2}

Because the system is rotationally symmetric, we can place point

P

on the positive x-axis at

(2 r, 0)

. This strategic choice simplifies the subsequent algebra significantly.

The Power of the Chord of Contact

We need to find the coordinates of

A

and

B

, the points where the tangents from

P

touch the inner circle. Instead of using slope-based methods, we utilize the Chord of Contact formula.

For any circle

x^{2} + y^{2} = r^{2}

and an external point

(x_{1}, y_{1})

, the line connecting the points of tangency is given by

x x_{1} + y y_{1} = r^{2}

Substituting

P (2 r, 0)

into the formula, we obtain:

x (2 r) + y (0) = r^{2}

This simplifies to

2 r x = r^{2}

, or:

x = \frac{r}{2}

The chord of contact

A B

is therefore a vertical line located at

x = \frac{r}{2}

Finding the Vertices

Since points

A

and

B

lie on the inner circle

C_{1}

and the line

x = \frac{r}{2}

, we substitute the line equation into the circle equation:

(\frac{r}{2})^{2} + y^{2} = r^{2}

Expanding and solving for

y

\frac{r ^{2}}{4} + y^{2} = r^{2} \Rightarrow y^{2} = \frac{3 r ^{2}}{4} \Rightarrow y = \pm \frac{r 3}{2}

Thus, the vertices of the triangle are

P (2 r, 0)

A (\frac{r}{2}, \frac{r 3}{2})

, and

B (\frac{r}{2}, - \frac{r 3}{2})

Final Calculation

The centroid

G (x_{G}, y_{G})

is the arithmetic mean of the vertices. Calculating the x-coordinate:

x_{G} = \frac{2 r + \frac{r}{2} + \frac{r}{2}}{3} = \frac{3 r}{3} = r

Calculating the y-coordinate:

y_{G} = \frac{0 + \frac{r 3}{2} - \frac{r 3}{2}}{3} = 0

The centroid

G

is located at

(r, 0)

. Substituting these coordinates into the equation of the inner circle

x^{2} + y^{2} = r^{2}

, we get

r^{2} + 0^{2} = r^{2}

The centroid $G$ lies exactly on the circumference of the inner circle.

Common Traps & FAQs

Wait, why can we just assume

P

is at

(2 r, 0)

? Isn't that a specific case?

That is a fantastic question! This is the beauty of coordinate geometry. Because the circles are concentric, the system is rotationally symmetric. Any point

P

C_{2}

is equivalent to any other point

P

C_{2}

relative to the center. By choosing

(2 r, 0)

, we aren't losing generality; we are simply choosing the most convenient frame of reference to make the algebra elegant. If you chose

(2 r cos θ, 2 r sin θ)

, you would arrive at the exact same result, just with more trigonometric clutter!

I always forget the Chord of Contact formula. Is there another way?

You could use the slope-point form of a line

y - y_{1} = m (x - x_{1})

and the condition of tangency

c^{2} = r^{2} (1 + m^{2})

, but that is a long, winding road. In JEE Advanced, time is currency. Memorizing

T = 0

is a high-yield investment. It works for any conic section, not just circles! It is the single most efficient tool for problems involving tangents from an external point.

Why does the centroid lie on the inner circle? Is there a physical intuition?

Think of it as a balance of forces. The symmetry of the tangents

P A

and

P B

forces the centroid to lie on the axis of symmetry (the x-axis in our setup). The distance calculation confirms that this balance point is exactly at a distance

r

from the origin, placing it perfectly on the circumference of

C_{1}

. It is a beautiful geometric harmony!

Animated Solution for Mathematics - Circles: $C_{1}$ and $C_{2}$ are two concentric circles, the radius of $C_{2}$ being twice that of $C_{1}$ . From a point $P$ on $C_{2}$ , tangents $P A$ and $P B$ are drawn to $C_{1}$ . Prove that the centroid of the triangle $P A B$ lies on $C_{1}$ .

Visualized Solution

Setting up the Circles $C_{1}$ and $C_{2}$

Defining Point $P$ on $C_{2}$

Drawing Tangents $P A$ and $P B$

The Chord of Contact $A B$

Equation of the Chord of Contact

Substituting Point $P$ into the Equation

Simplifying the Chord Equation

Finding the y-coordinates of $A$ and $B$

Solving for $y$

Coordinates of $A$ and $B$

Formula for the Centroid $G$

Calculating the x-coordinate of $G$

Calculating the y-coordinate of $G$

Final Verification

The Sigma Insight: Equation of Tangent and Normal

Analyzing the Setup

The Power of the Chord of Contact

Finding the Vertices

Final Calculation

Similar Questions

Circles with radii 3, 4 and 5 touch each other externally. If $P$ is the point of intersection of tangents to these circles at their points of contact, find the distance of $P$ from the points of contact.

Two circles each of radius 5 units touch each other at the point $(1, 2)$ . If the equation of their common tangent is $4 x + 3 y = 10$ , and $C_{1} (α, β)$ and $C_{2} (γ, δ)$ , $C_{1} \neq = C_{2}$ are their centres, then $∣ (α + β) (γ + δ) ∣$ is equal to .

If a tangent to the circle $x^{2} + y^{2} = 1$ intersects the coordinate axes at distinct points $P$ and $Q$ , then the locus of the mid-point of $P Q$ is

Visualized Solution

Setting up the Circles .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } C1​ and .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } C2​

Defining Point .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } P on .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } C2​

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

The Chord of Contact .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } AB

Equation of the Chord of Contact

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

Simplifying the Chord Equation

Finding the y-coordinates of .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } A and .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } B

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

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

Formula for the Centroid .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } G

Calculating the x-coordinate of .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } G

Calculating the y-coordinate of .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } G

Final Verification

The Sigma Insight: Equation of Tangent and Normal

Analyzing the Setup

The Power of the Chord of Contact

Finding the Vertices

Final Calculation

Similar Questions

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If a tangent to the circle x2+y2=1 intersects the coordinate axes at distinct points P and Q, then the locus of the mid-point of PQ is

Setting up the Circles $C_{1}$ and $C_{2}$

Defining Point $P$ on $C_{2}$

Drawing Tangents $P A$ and $P B$

The Chord of Contact $A B$

Substituting Point $P$ into the Equation

Finding the y-coordinates of $A$ and $B$

Solving for $y$

Coordinates of $A$ and $B$

Formula for the Centroid $G$

Calculating the x-coordinate of $G$

Calculating the y-coordinate of $G$

If a tangent to the circle $x^{2} + y^{2} = 1$ intersects the coordinate axes at distinct points $P$ and $Q$ , then the locus of the mid-point of $P Q$ is