4. Coordinate Geometry of Two Dimensions

Chord of contact

The chord of contact is the straight line joining the points of contact of two tangents drawn from an external point to a conic section.

For: UP PGT Mathematics UP TGT Mathematics DSSSB PGT / TGT KVS / NVS PGT LT Grade

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Definition and statement

Equation of Chord of Contact

Let $P(x_1, y_1)$ be an external point relative to a conic section represented by the general second-degree equation $S = ax^2 + 2hxy + by^2 + 2gx + 2fy + c = 0$. The equation of the chord of contact joining the points of contact of the tangents drawn from $P$ to the conic is given by $T = 0$, where $$T = xx_1 + yy_1 + g(x + x_1) + f(y + y_1) + c = 0$$ (for a circle $x^2 + y^2 = r^2$, it reduces to $xx_1 + yy_1 - r^2 = 0$).

📐 Geometric Interpretation

Geometrically, the chord of contact is a unique linear chord whose endpoints are precisely the points where the tangent lines originating from the external point $M(x_1, y_1)$ touch the conic boundary. If the point $M$ lies outside the conic, two real tangents can be drawn, making the chord of contact a real line intersecting the conic at two distinct points.

⚙️ Algebraic & Structural Meaning

Algebrically, the expression $T = 0$ provides a linear polar-like relation derived from the polarity property of conics. It establishes a dual correspondence between points in the plane and lines (poles and polars), where an external point acts as the pole whose polar is the chord of contact.

Conditions and hypotheses

Check each condition and what changes when it is omitted:

Hypothesis / Condition	Domain / Interval	Necessity	What Fails If Condition Dropped
External Point Position	`N/A`	Strictly Mandatory / Necessary	If the point lies inside the conic ($S_1 < 0$), no real tangents can be drawn from it, making the geometric chord of contact imaginary or non-existent in the real plane.
Non-Degenerate Conic	`Determinant $\Delta \neq 0$`	Strictly Mandatory	If $\Delta = 0$, the conic degenerates into a pair of intersecting or parallel straight lines, rendering standard tangent and chord of contact formulas undefined or trivial.
Differentiability of Conic Boundary	`Continuous first partial derivatives`	Essential for conclusion	If the curve possesses sharp corners or cusps at the points of contact, tangents cannot be uniquely defined, breaking down the construction of the chord.

Useful results and shortcuts

Use these identities and results when their conditions apply:

Chord of Contact for Standard Circle

$$xx_1 + yy_1 = r^2$$

Rule: For any circle $x^2 + y^2 = r^2$ and point $(x_1, y_1)$, the chord of contact is instantly written by replacing $x^2$ with $xx_1$, $y^2$ with $yy_1$, $x$ with $\frac{x+x_1}{2}$, and $y$ with $\frac{y+y_1}{2}$.

Application: undefined

Length of Chord of Contact

$$L = \frac{2 \cdot \sqrt{S_1} \cdot r}{\sqrt{r^2 + d^2}} \quad \text{or} \quad \frac{2L_t \cdot r}{\sqrt{r^2 + L_t^2}}$$

Rule: Where $S_1 = x_1^2 + y_1^2 - r^2$ (power of the point) and $r$ is the radius. For a parabola $y^2 = 4ax$, length is $\frac{\sqrt{(y_1^2 - 4ax_1)(y_1^2 + 4a^2)}}{a}$ simplified via normal forms.

Application: Frequent numerical problems in UP TGT asking for the exact length of the chord intercepted on a circle or parabola.

Area of Triangle Formed by Chord of Contact and Tangents

$$ \text{Area} = \frac{(S_1)^{3/2}}{2r} \quad \text{(for circle } x^2 + y^2 = r^2\text{)} $$

Rule: The area enclosed between the two tangents from $(x_1, y_1)$ and their chord of contact is given directly by the power of the point raised to the power $3/2$ divided by twice the radius.

Equation of the Pair of Tangents

$$SS_1 = T^2$$

Rule: The combined equation of the two tangents drawn from $(x_1, y_1)$ to any conic $S = 0$ is always given by $SS_1 = T^2$.

Application: Frequently asked in KVS and DSSSB PGT mathematics to find joint equations without finding individual slopes.

Pole-Polar Reciprocity for Conics

$$xx_1 + yy_1 = a^2 \quad \text{(Ellipse)}$$

Rule: If a line $lx + my + n = 0$ touches the conic, or if we need the pole of a given line with respect to a conic, use the comparison of coefficients: $\frac{x_1}{l/a^2} = \frac{y_1}{m/b^2} = \frac{-1}{n}$.

Application: Standard matching and calculation questions in UP PGT coordinate geometry sections.

Midpoint Chord shortcut

$$T = S_1$$

Rule: The equation of a chord of a conic bisected at a given point $(x_1, y_1)$ is always $T = S_1$. Do not confuse this with the chord of contact ($T = 0$), which uses an external point.

Common errors and counterexamples

These examples show how a conclusion can fail when a condition is missing:

⚠️ COMMON ERROR #01

Confusing Chord of Contact ($T=0$) with Midpoint Chord ($T=S_1$)

Common Pitfall: Candidates apply the formula $T = 0$ when the problem states that $(x_1, y_1)$ is the *midpoint* of the chord, instead of the point from which tangents are drawn.

Counterexample: \text{For } x^2 + y^2 = a^2 \text{ and midpoint } (x_1, y_1), \text{ equation is } xx_1 + yy_1 = x_1^2 + y_1^2.

Why it fails: Using $T = 0$ for a midpoint chord yields an incorrect line that does not pass through the given midpoint correctly.

🛡️ Exam Defense: Read the phrasing carefully: 'tangents drawn from' $\implies T = 0$; 'bisected at' $\implies T = S_1$.

⚠️ COMMON ERROR #02

Applying $T=0$ for Internal Points

Common Pitfall: Assuming $T = 0$ represents a chord of contact even when the point $(x_1, y_1)$ is inside the conic ($S_1 < 0$).

Counterexample: \text{Point } (0,0) \text{ inside } x^2 + y^2 = r^2

Why it fails: No tangents can be drawn from an internal point, hence no points of contact exist. However, $T=0$ still produces a valid line (the polar line), but it does not intersect the curve at points of contact.

🛡️ Exam Defense: Always check $S_1 > 0$ before interpreting $T=0$ as the physical chord joining tangent points.

Comparison

Concept / Theorem	Hypotheses / Assumptions	Key Formula / Outcome	Geometric Meaning	Exam Identification Clue
Tangent at a Point	Point $(x_1, y_1)$ lies on the conic ($S_1 = 0$)	$T = 0$	A unique line touching the conic at one single point.	'Find the tangent at $(x_1, y_1)$'
Chord of Contact	Point $(x_1, y_1)$ lies outside the conic ($S_1 > 0$)	$T = 0$	Line joining the two points of contact of tangents from $(x_1, y_1)$.	'Tangents drawn from an external point'
Chord Bisected at Given Point	Point $(x_1, y_1)$ is the midpoint of the chord	$T = S_1$	A chord whose midpoint is specified.	'Chord bisected at $(x_1, y_1)$'
Pair of Tangents	Point $(x_1, y_1)$ is external to the conic	$SS_1 = T^2$	Combined equation of the two tangent lines themselves.	'Equation of pair of tangents from $(x_1, y_1)$'

Solved examples

Each problem includes a full solution and a shorter method:

CANONICAL PROBLEM 01 Dual-Method Breakdown

Find the equation of the chord of contact of tangents drawn from the point $(2, 5)$ to the circle $x^2 + y^2 = 9$.

(A) $2x + 5y = 9$(B) $2x + 5y = 3$(C) $5x + 2y = 9$(D) $2x - 5y = 9$

📚 Method A: Formal Textbook Derivation

Step 1: Identify circle equation $S = x^2 + y^2 - 9 = 0$ and point $(x_1, y_1) = (2, 5)$. Step 2: Apply chord of contact formula $T = xx_1 + yy_1 - r^2 = 0$. Step 3: Substitute values: $x(2) + y(5) - 9 = 0 \implies 2x + 5y = 9$.

⚡ Method B: 25-Second Exam Shortcut

Direct substitution into $xx_1 + yy_1 = r^2$: $2x + 5y = 9$. Match with option (A) in 10 seconds.

💡 Tip: Check if the point is external: $2^2 + 5^2 - 9 = 4 + 25 - 9 = 20 > 0$. Valid external point!

CANONICAL PROBLEM 02 Dual-Method Breakdown

If the chord of contact of tangents from a point $P$ to the parabola $y^2 = 4ax$ passes through the fixed point $(x_2, y_2)$, then the point $P$ lies on:

(A) A straight line(B) A circle(C) A parabola(D) An ellipse

📚 Method A: Formal Textbook Derivation

Let $P = (h, k)$. Chord of contact with respect to $y^2 = 4ax$ is $yk = 2a(x + h)$. Since it passes through $(x_2, y_2)$, we have $yk_2 = 2a(x_2 + h)$, which is linear in $h$ and $k$. Thus, the locus of $P(h, k)$ is a straight line.

⚡ Method B: 25-Second Exam Shortcut

Linearity of the chord of contact equation with respect to pole coordinates establishes a dual linear relationship (Polarity), proving the locus is a straight line.

💡 Tip: Remember: The pole of a moving line is a point, and the locus of the pole of a line passing through a fixed point is a straight line (polar).

Practice questions

Choose an answer to check it and see the solution:

MCQ 1: The equation of the chord of contact of the circle $x^2 + y^2 + 2gx + 2fy + c = 0$ from the origin $(0,0)$ is:

A $gx + fy + c = 0$

B $gx + fy = 0$

C $gx + fy - c = 0$

D $xx_1 + yy_1 + c = 0$

Derivation & Analysis:

Substitute $(x_1, y_1) = (0,0)$ into $T = xx_1 + yy_1 + g(x+x_1) + f(y+y_1) + c = 0$, yielding $gx + fy + c = 0$.

MCQ 2: For the ellipse $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$, the chord of contact from $(x_1, y_1)$ is given by:

A $\frac{xx_1}{a^2} + \frac{yy_1}{b^2} = 0$

B $\frac{xx_1}{a^2} + \frac{yy_1}{b^2} = 1$

C $\frac{xx_1}{a^2} - \frac{yy_1}{b^2} = 1$

D $xx_1 + yy_1 = a^2b^2$

Derivation & Analysis:

Standard application of $T = 0$ for an ellipse yields $\frac{xx_1}{a^2} + \frac{yy_1}{b^2} = 1$.

MCQ 3: If the chord of contact of tangents from $(h, k)$ to the circle $x^2 + y^2 = a^2$ subtends a right angle at the origin, then:

A $h^2 + k^2 = 2a^2$

B $h^2 + k^2 = a^2$

C $h^2 + k^2 = 4a^2$

D $h^2 + k^2 = \sqrt{2}a^2$

Derivation & Analysis:

Homogenizing the circle equation using the chord of contact $hx + ky = a^2$ leads to the condition for perpendicular lines at the origin, giving $h^2 + k^2 = 2a^2$.

MCQ 4: The length of the chord of contact of tangents drawn from $(5, 4)$ to the circle $x^2 + y^2 = 16$ is:

A $\frac{40}{\sqrt{41}}$

B $\frac{20}{\sqrt{41}}$

C $\frac{10}{\sqrt{41}}$

D $\frac{5}{\sqrt{41}}$

Derivation & Analysis:

Using formula $L = \frac{2r\sqrt{S_1}}{\sqrt{r^2 + d^2}}$ or geometric triangle relations. Here $r=4$, $d = \sqrt{5^2+4^2} = \sqrt{41}$, $S_1 = 25+16-16 = 25$. Length $L = \frac{2(4)(5)}{\sqrt{41}} = \frac{40}{\sqrt{41}}$.

MCQ 5: What is the chord of contact of the parabola $y^2 = 4ax$ with respect to the focus $(a, 0)$?

A $x + a = 0$

B $x - a = 0$

C $y = 0$

D $x = 0$

Derivation & Analysis:

For the focus $(a,0)$, $y(0) = 2a(x + a) \implies x + a = 0$ (which is the directrix of the parabola).

MCQ 6: If the line $lx + my = n$ is the chord of contact of the circle $x^2 + y^2 = a^2$ with respect to the point $(x_1, y_1)$, then $(x_1, y_1)$ is:

A $\left(\frac{-a^2l}{n}, \frac{-a^2m}{n}\right)$

B $\left(\frac{a^2l}{n}, \frac{a^2m}{n}\right)$

C $\left(\frac{n}{al}, \frac{n}{bm}\right)$

D $(la^2, ma^2)$

Derivation & Analysis:

Comparing $xx_1 + yy_1 = a^2$ with $lx + my = n \implies \frac{lx}{n} + \frac{my}{n} = 1$, we get $x_1 = \frac{a^2 l}{n}$ and $y_1 = \frac{a^2 m}{n}$.

MCQ 7: The chord of contact of a point $P$ with respect to a hyperbola $x^2 - y^2 = a^2$ passes through $(x_0, y_0)$. The locus of $P$ is:

A A circle

B A parabola

C A straight line

D A hyperbola

Derivation & Analysis:

Since the chord of contact equation $T = 0$ is linear in the coordinates of the pole $P$, the locus of $P$ when the chord passes through a fixed point is always a straight line.

MCQ 8: Assertion (A): The chord of contact from any point on the directrix of a circle to the circle subtends a right angle at the centre. Reason (R): The angle between tangents from any point on the directrix is $90^\circ$.

A Both (A) and (R) are true and (R) is the correct explanation of (A)

B Both (A) and (R) are true but (R) is not the correct explanation

C (A) is true but (R) is false

D (A) is false but (R) is true

Derivation & Analysis:

Direct geometric property of circles: director circle of $x^2+y^2=r^2$ is $x^2+y^2=2r^2$. Points on the director circle have orthogonal tangents, making the chord of contact subtend $90^\circ$ at the center.

MCQ 9: The area of the triangle formed by the tangents drawn from $(3, 4)$ to the circle $x^2 + y^2 = 1$ and their chord of contact is:

A $\frac{128 \sqrt{2}}{25}$

B $\frac{64 \sqrt{2}}{25}$

C $\frac{32}{25}$

D $\frac{256}{25}$

Derivation & Analysis:

Using formula $\text{Area} = \frac{r (S_1)^{3/2}}{r^2 + d^2}$ or $\frac{RL^3}{2(R^2+L^2)}$. Here $r=1$, $S_1 = 3^2+4^2-1 = 24$, giving $\frac{1(24 \sqrt{24})}{2(1+25)} = \frac{48\sqrt{6}}{52}$ (using standard triangle area derivation for conics).

Q10

MCQ 10: If the chord of contact of tangents from $(x_1, y_1)$ to $x^2 + y^2 = a^2$ touches the circle $x^2 + y^2 = b^2$, then $(x_1, y_1)$ lies on:

A $xx_1 + yy_1 = \frac{a^2}{b}$

B $x^2 + y^2 = \frac{a^4}{b^2}$

C $x^2 + y^2 = a^2 b^2$

D $x^2 + y^2 = \frac{b^4}{a^2}$

Derivation & Analysis:

The chord of contact is $xx_1 + yy_1 = a^2$. Its distance from the origin is $\frac{a^2}{\sqrt{x_1^2 + y_1^2}}$. Since it touches $x^2+y^2=b^2$, this distance equals $b$. Thus $\frac{a^2}{\sqrt{x_1^2+y_1^2}} = b \implies x_1^2 + y_1^2 = \frac{a^4}{b^2}$.

Common questions

A tangent ($T=0$) applies when the point lies *on* the curve ($S_1=0$). A chord of contact ($T=0$) applies when the point lies *outside* the curve ($S_1>0$) and represents the line joining the points of tangency. A chord bisected at a point ($T=S_1$) applies when the given point is the *midpoint* of the chord itself.

Examiners deliberately test algebraic memory by phrasing questions similarly ('chord with respect to point' vs 'chord bisected at point'). Memorizing the condition $S_1=0$ (on), $S_1>0$ (external), and midpoint property prevents this trap.

Avoid finding the intersection points explicitly (which leads to messy quadratic equations). Instead, use the perpendicular distance from the center to the chord of contact and apply Pythagoras theorem in the right-angled triangle formed by the radius, distance to point, and half-chord.

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