JEE Advanced 1997

LEVELJEE Main

Animated Solution for Mathematics - Circles: The chords of contact of the pair of tangents drawn from each point on the line $2 x + y = 4$ to circle $x^{2} + y^{2} = 1$ pass through the point .........

Q: Wait, why do we use the $T=0$ formula for the chord of contact?

The $T=0$ notation is a powerful shorthand in coordinate geometry. For a circle $x^2 + y^2 = r^2$, the equation of the chord of contact from an external point $(x_1, y_1)$ is indeed $xx_1 + yy_1 = r^2$. It is the most direct way to define the line connecting the two points of tangency without having to find the coordinates of those points themselves.

Q: How do I know if the point $(1/2, 1/4)$ is the correct fixed point?

Once you have the equation in the form $L_1 + hL_2 = 0$, the fixed point must satisfy both $L_1 = 0$ and $L_2 = 0$ simultaneously. Since the equation must hold for any value of $h$, the only way for the equation to be independent of $h$ is if the coefficient of $h$ is zero. Solving those two linear equations gives you the unique point through which all such chords must pass.

Visualized Solution

Visualizing the Setup

Circle: $x^{2} + y^{2} = 1$
Line: $2 x + y = 4$
Objective: Find the fixed point through which all chords of contact pass.

General Point $P (h, k)$

Let point $P (h, k)$ lie on the line $2 x + y = 4$ .
Substitution: $2 h + k = 4 ⟹ k = 4 - 2 h$ .

Drawing the Chord of Contact

Draw tangents from $P (h, k)$ to the circle.
The line joining the points of tangency is the Chord of Contact.

Equation of Chord of Contact

For a circle $x^{2} + y^{2} = a^{2}$ , the chord of contact from $P (h, k)$ is given by $T = 0$ .
Result: $h x + k y = 1$

Substituting the Constraint

Substitute $k = 4 - 2 h$ into the chord equation.
Raw Equation: $h x + (4 - 2 h) y = 1$

Expanding the Equation

Expand the brackets: $h x + 4 y - 2 h y = 1$
Rearrange to group terms: $4 y - 1 + h x - 2 h y = 0$

Family of Lines $L_{1} + λ L_{2} = 0$

Factor out $h$ : $(4 y - 1) + h (x - 2 y) = 0$
This matches the Family of Lines form $L_{1} + λ L_{2} = 0$ .

Another Point on the Line

If we choose a different point $P^{'} (h^{'}, k^{'})$ on the line, we get a different chord of contact.
All such chords belong to the same family of lines.

Finding the Fixed Point

To find the intersection, solve $L_{1} = 0$ and $L_{2} = 0$ .
1) $4 y - 1 = 0 ⟹ y = \frac{1}{4}$

Solving for $x$

2) $x - 2 y = 0 ⟹ x = 2 y$
Substitute $y = \frac{1}{4}$ : $x = 2 (\frac{1}{4}) = \frac{1}{2}$

Conclusion

The fixed point is $(\frac{1}{2}, \frac{1}{4})$ .
All chords of contact from points on $2 x + y = 4$ pass through this point.

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The Sigma Insight: Length of Tangent and Chord of Contact

Solution Diagram

Analyzing the Setup

Imagine you are standing on the coordinate plane, looking at a unit circle centered at the origin, defined by the equation

x^{2} + y^{2} = 1

. A line

2 x + y = 4

is positioned nearby.

For any point

P (h, k)

chosen on this line, we draw two tangents from

P

to the circle. Connecting the points of tangency forms a chord of contact.

The objective is to find the fixed point through which all such chords of contact must pass, regardless of the position of

P

on the line.

The Power of the General Point

We define our point

P (h, k)

on the line

2 x + y = 4

. Because

P

is constrained to this line, its coordinates must satisfy the equation:

2 h + k = 4

This allows us to express

k

k = 4 - 2 h

. By reducing the complexity from two variables to one, we simplify the problem significantly.

This is a classic JEE strategy: whenever a point is constrained to a line, use the line's equation to reduce the number of variables.

The Equation of the Chord

For any point

(h, k)

outside the circle

x^{2} + y^{2} = 1

, the equation of the chord of contact is given by the

T = 0

form:

h x + k y = 1

Substituting our constraint

k = 4 - 2 h

into this equation, the chord of contact becomes:

h x + (4 - 2 h) y = 1

The Dance of the Family of Lines

To isolate the variable

h

, we expand the equation:

h x + 4 y - 2 h y = 1

Rearranging the terms to group those containing

h

together, we obtain:

(4 y - 1) + h (x - 2 y) = 0

This equation is in the form

L_{1} + h L_{2} = 0

, where

L_{1} = 4 y - 1

and

L_{2} = x - 2 y

. This represents a family of lines passing through the intersection of

L_{1} = 0

and

L_{2} = 0

The Final Intersection

To find the fixed point, we solve the system of equations formed by

L_{1}

and

L_{2}

4 y - 1 = 0 ⟹ y = \frac{1}{4}

x - 2 y = 0 ⟹ x = 2 y

Substituting

y = \frac{1}{4}

into the second equation, we find:

x = 2 (\frac{1}{4}) = \frac{1}{2}

Thus, the fixed point is $(\frac{1}{2}, \frac{1}{4})$ . No matter how

P

moves along the line

2 x + y = 4

, the chord of contact is destined to pivot around this single, elegant coordinate.

Common Traps & FAQs

Wait, why do we use the

T = 0

formula for the chord of contact?

The

T = 0

notation is a powerful shorthand in coordinate geometry. For a circle

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

, the equation of the chord of contact from an external point

(x_{1}, y_{1})

is indeed

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

. It is the most direct way to define the line connecting the two points of tangency without having to find the coordinates of those points themselves.

Can we just pick two points on the line

2 x + y = 4

and find the intersection of their chords?

Absolutely! That is a valid strategy. If you pick two points, say

(2, 0)

and

(0, 4)

, you can find the chord of contact for each and solve for their intersection. However, the 'family of lines' method is more elegant because it proves the property for *all* points on the line simultaneously, rather than just two specific cases.

How do I know if the point

(1/2, 1/4)

is the correct fixed point?

Once you have the equation in the form

L_{1} + h L_{2} = 0

, the fixed point must satisfy both

L_{1} = 0

and

L_{2} = 0

simultaneously. Since the equation must hold for any value of

h

, the only way for the equation to be independent of

h

is if the coefficient of

h

is zero. Solving those two linear equations gives you the unique point through which all such chords must pass.

Animated Solution for Mathematics - Circles: The chords of contact of the pair of tangents drawn from each point on the line $2 x + y = 4$ to circle $x^{2} + y^{2} = 1$ pass through the point .........

Visualized Solution

Visualizing the Setup

General Point $P (h, k)$

Drawing the Chord of Contact

Equation of Chord of Contact

Substituting the Constraint

Expanding the Equation

Family of Lines $L_{1} + λ L_{2} = 0$

Another Point on the Line

Finding the Fixed Point

Solving for $x$

Conclusion

The Sigma Insight: Length of Tangent and Chord of Contact

Analyzing the Setup

The Power of the General Point

The Equation of the Chord

The Dance of the Family of Lines

The Final Intersection

Similar Questions

Comprehension Passage

Let P be a point on the circle S with both coordinates being positive. Let the tangent to S at P intersect the coordinate axes at the points M and N. Then, the mid-point of the line segment MN must lie on the curve

The locus of the mid-point of the chord of contact of tangents drawn from points lying on the straight line $4 x - 5 y = 20$ to the circle $x^{2} + y^{2} = 9$ is

If the circle $C_{1} : x^{2} + y^{2} = 16$ intersects another circle $C_{2}$ of radius 5 in such a manner that common chord is of maximum length and has a slope equal to $3/4$ , then the coordinates of the centre of $C_{2}$ are .........

For the circle $x^{2} + y^{2} = r^{2}$ , find the value of $r$ for which the area enclosed by the tangents drawn from the point $P (6, 8)$ to the circle and the chord of contact is maximum.

Let the latus rectum of the parabola $y^{2} = 4 x$ be the common chord to the circles $C_{1}$ and $C_{2}$ each of them having radius $25$ . Then, the distance between the centres of the circles $C_{1}$ and $C_{2}$ is:

A circle with centre $(2, 3)$ and radius 4 intersects the line $x + y = 3$ at the points $P$ and $Q$ . If the tangents at $P$ and $Q$ intersect at the point $S (α, β)$ , then $4 α - 7 β$ is equal to

Let $B$ be the centre of the circle $x^{2} + y^{2} - 2 x + 4 y + 1 = 0$ . Let the tangents at two points $P$ and $Q$ on the circle intersect at the point $A (3, 1)$ . Then $8 \cdot (\frac{area Δ A P Q}{area Δ B P Q})$ is equal to .

The area of the triangle formed by the tangents from the point $(4, 3)$ to the the circle $x^{2} + y^{2} = 9$ and the line joining their points of contact is .........

Let $x^{2} + y^{2} - 4 x - 2 y - 11 = 0$ be a circle. A pair of tangents from the point $(4, 5)$ with a pair of radii form a quadrilateral of area .........

If the tangents at the points $P$ and $Q$ on the circle $x^{2} + y^{2} - 2 x + y = 5$ meet at the point $R (\frac{9}{4}, 2)$ , then the area of the triangle $P QR$ is

Visualized Solution

Visualizing the Setup

General Point .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } P(h,k)

Drawing the Chord of Contact

Equation of Chord of Contact

Substituting the Constraint

Expanding the Equation

Family of Lines .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } L1​+λL2​=0

Another Point on the Line

Finding the Fixed Point

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

Conclusion

The Sigma Insight: Length of Tangent and Chord of Contact

Analyzing the Setup

The Power of the General Point

The Equation of the Chord

The Dance of the Family of Lines

The Final Intersection

Similar Questions

Comprehension Passage

Let P be a point on the circle S with both coordinates being positive. Let the tangent to S at P intersect the coordinate axes at the points M and N. Then, the mid-point of the line segment MN must lie on the curve

.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 the chord of contact of tangents drawn from points lying on the straight line 4x−5y=20 to the circle x2+y2=9 is

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } For the circle x2+y2=r2, find the value of r for which the area enclosed by the tangents drawn from the point P(6,8) to the circle and the chord of contact is maximum.

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The area of the triangle formed by the tangents from the point (4,3) to the the circle x2+y2=9 and the line joining their points of contact is .........

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Let x2+y2−4x−2y−11=0 be a circle. A pair of tangents from the point (4,5) with a pair of radii form a quadrilateral of area .........

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If the tangents at the points P and Q on the circle x2+y2−2x+y=5 meet at the point R(49​,2), then the area of the triangle PQR is

General Point $P (h, k)$

Family of Lines $L_{1} + λ L_{2} = 0$

Solving for $x$

The locus of the mid-point of the chord of contact of tangents drawn from points lying on the straight line $4 x - 5 y = 20$ to the circle $x^{2} + y^{2} = 9$ is

For the circle $x^{2} + y^{2} = r^{2}$ , find the value of $r$ for which the area enclosed by the tangents drawn from the point $P (6, 8)$ to the circle and the chord of contact is maximum.

The area of the triangle formed by the tangents from the point $(4, 3)$ to the the circle $x^{2} + y^{2} = 9$ and the line joining their points of contact is .........

Let $x^{2} + y^{2} - 4 x - 2 y - 11 = 0$ be a circle. A pair of tangents from the point $(4, 5)$ with a pair of radii form a quadrilateral of area .........

If the tangents at the points $P$ and $Q$ on the circle $x^{2} + y^{2} - 2 x + y = 5$ meet at the point $R (\frac{9}{4}, 2)$ , then the area of the triangle $P QR$ is