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4. Coordinate Geometry of Two Dimensions

Common tangents to two conics

This section covers the determination and equations of common tangents to two conics (such as circles, parabolas, ellipses, and hyperbolas).

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

Condition of Co-tangency & Pole-Polar Tangency Principle
Let two conics be given by equations $S_1(x, y) = 0$ and $S_2(x, y) = 0$. A straight line $y = mx + c$ is a common tangent to both conics if and only if the length of the perpendicular from the respective centers (or foci/vertices) equals the corresponding radius (or semi-axis parameters), or equivalently, the discriminant obtained by substituting the line equation into each conic yields equal independent parameter conditions for $c$ in terms of $m$.

📐 Geometric Interpretation

Common tangents divide into direct (external) tangents and transverse (internal) tangents. For two non-intersecting or exterior conics, exactly four common tangents exist: two direct and two transverse, passing through the external and internal centers of similitude respectively.

⚙️ Algebraic & Structural Meaning

Translates geometrical tangency into algebraic consistency of quadratic equations in one variable. By enforcing the condition that the discriminant $\Delta = 0$ for the intersection of $y = mx + c$ with both conics, we obtain a relation that explicitly solves for slope $m$ and intercept $c$.

Conditions and hypotheses

Check each condition and what changes when it is omitted:

Hypothesis / Condition Domain / Interval Necessity What Fails If Condition Dropped
Non-Degeneracy of Conics Discriminant $\Delta \neq 0$ for individual quadratic forms Strictly Mandatory If a conic degenerates into a pair of intersecting or coincident lines, the standard tangent formulas fail or yield redundant/incorrect line pairs.
Real Domain for Tangent Slopes Slope parameter $m \in \mathbb{R}$ Mandatory If the discriminant for determining $m$ yields purely imaginary roots, no real common tangents exist (conics are positioned such that one lies entirely inside the other without intersection).
Relative Position & Separation Distance between centers $d > r_1 + r_2$ (for circles) Essential for determining tangent count If conics intersect or overlap, the number of common tangents reduces from 4 to 3, 2, 1, or 0.

Useful results and shortcuts

Use these identities and results when their conditions apply:

01

Condition for a line to touch two circles

$$c^2 = a^2(1 + m^2) \quad \text{and} \quad (mx_1 - y_1 + c)^2 = r_2^2(1 + m^2)$$

Rule: Equating the intercept $c$ derived from the standard circle $x^2 + y^2 = a^2$ to the circle $(x - x_1)^2 + (y - y_1)^2 = r_2^2$ provides a rapid quadratic in $m$.

Application: Frequently asked in UP PGT to find the number or equation of common tangents between two circles of unequal radii.
02

Number of Common Tangents Based on Centers and Radii

$$ \begin{aligned} d > r_1 + r_2 & \implies 4 \text{ tangents} \\ d = r_1 + r_2 & \implies 3 \text{ tangents} \\ |r_1 - r_2| < d < r_1 + r_2 & \implies 2 \text{ tangents} \\ d = |r_1 - r_2| & \implies 1 \text{ tangent} \\ d < |r_1 - r_2| & \implies 0 \text{ tangents} \end{aligned} $$

Rule: Instantly determine the exact count of common tangents by comparing the distance between centers $d$ with the sum and difference of radii ($r_1, r_2$).

Application: Standard 1-mark theoretical question in UP TGT and KVS PGT.
03

Direct Tangents via Center of Similitude

$$ x_{sim} = \frac{r_1 x_2 - r_2 x_1}{r_1 - r_2}, \quad y_{sim} = \frac{r_1 y_2 - r_2 y_1}{r_1 - r_2} $$

Rule: The external center of similitude divides the join of centers externally in the ratio of their radii. Direct common tangents pass through this point.

Application: Crucial for writing equations of direct common tangents without computing slopes.
04

Transverse Tangents Similitude Point

$$ x_{trans} = \frac{r_1 x_2 + r_2 x_1}{r_1 + r_2}, \quad y_{trans} = \frac{r_1 y_2 + r_2 y_1}{r_1 + r_2} $$

Rule: The internal center of similitude divides the join of centers internally in the ratio of radii $r_1 : r_2$. Transverse common tangents pass through this point.

Application: Directly tested in DSSSB PGT Mathematics for finding internal tangents.
05

Common Tangent to Parabola and Circle / Ellipse

$$ y = mx + \frac{a}{m} \implies \text{ substitute into conic and set } \Delta = 0 $$

Rule: Use the slope-form tangent equation for one conic and enforce the tangency condition on the second conic to solve for $m$.

Application: Appears in UP PGT when mixing parabolas ($y^2 = 4ax$) and circles ($x^2 + y^2 = r^2$).
06

Rapid Elimination via Slope Verification

$$ m = \pm \sqrt{\frac{d^2 - (r_1 \mp r_2)^2}{(r_1 \mp r_2)^2} \text{ (geometric variants)}} $$

Rule: Check the slope options given in the exam against the perpendicular distance formula from the center to eliminate incorrect option pairs instantly.

Common errors and counterexamples

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

⚠️ COMMON ERROR #01

Confusing Internal and External Centers of Similitude

Common Pitfall: Using internal division formula for direct tangents or external division for transverse tangents.

Counterexample: Circles $C_1: x^2+y^2=1$ and $C_2: (x-5)^2+y^2=4$. Radii $r_1=1, r_2=2$.

Why it fails: External center divides externally (minus sign), internal divides internally (plus sign). Swapping them yields equations of secants or non-tangent lines.

🛡️ Exam Defense: Remember: Direct tangents $\implies$ External division ($r_1 - r_2$). Transverse tangents $\implies$ Internal division ($r_1 + r_2$).
⚠️ COMMON ERROR #02

Ignoring Sign Ambiguity in Intercept $c$

Common Pitfall: Dropping the $\pm$ sign when equating $c = \pm r \sqrt{1 + m^2}$, missing half of the common tangents.

Counterexample: For $y = mx + c$ touching $x^2 + y^2 = r^2$, $c = \pm r\sqrt{1+m^2}$.

Why it fails: Conics are symmetric about their centers; tangents always exist in parallel pairs with opposite y-intercept signs.

🛡️ Exam Defense: Always account for both positive and parallel negative intercept branches.

Comparison

Concept / TheoremHypotheses / AssumptionsKey Formula / OutcomeGeometric MeaningExam Identification Clue
Direct Common Tangents$d > r_1 + r_2$ or $d > |r_1 - r_2|$Passes through external similitude centerTangents that do not cross the line of centersOptions with same slope, outer touching
Transverse Common Tangents$d > r_1 + r_2$Passes through internal similitude centerTangents that cross the line of centers between circlesOptions with intersecting tangents at internal center
Orthogonal Conics / IntersectionTangents at intersection are perpendicular$g_1 g_2 + f_1 f_2 = c_1 + c_2$Angle between tangents is $90^\circ$Keywords: 'orthogonal circles' or 'cut at right angles'
Co-axial System TangentsFamily of circles sharing radical axisRadical center propertyTangents from radical center to all circles are equal in lengthKeywords: 'system of circles', 'common radical axis'

Solved examples

Each problem includes a full solution and a shorter method:

CANONICAL PROBLEM 01 Dual-Method Breakdown

Find the number of common tangents to the circles $x^2 + y^2 = 4$ and $x^2 + y^2 - 6x - 8y + 9 = 0$.

(A) 1(B) 2(C) 3(D) 4
✓ Correct Answer: (C)
📚 Method A: Formal Textbook Derivation
Step 1: Identify centers and radii. Circle 1: $C_1(0,0), r_1 = 2$. Circle 2: Center $C_2(3,4), r_2 = \sqrt{3^2 + 4^2 - 9} = \sqrt{16} = 4$. Step 2: Calculate distance between centers $d = \sqrt{(3-0)^2 + (4-0)^2} = 5$. Step 3: Compare $d$ with $r_1 + r_2$ and $|r_1 - r_2|$. Here $r_1 + r_2 = 2 + 4 = 6$ and $|r_1 - r_2| = |2 - 4| = 2$. Since $2 < 5 < 6$ (i.e., $|r_1 - r_2| < d < r_1 + r_2$), the circles intersect at two distinct points. Step 4: Intersecting circles possess exactly 3 common tangents.
⚡ Method B: 25-Second Exam Shortcut
Compute $d = 5$. Radii are 2 and 4. Sum = 6, Diff = 2. Since $2 < 5 < 6$, circles intersect $\implies$ exactly 3 common tangents.
💡 Tip: Memorize the 5 positional inequalities for circle distances to answer.
CANONICAL PROBLEM 02 Dual-Method Breakdown

The equation of the direct common tangents to the circles $x^2 + y^2 = 1$ and $x^2 + y^2 - 6x - 8y + 24 = 0$ is:

(A) $3x + 4y - 5 = 0$(B) $3x - 4y = 5$(C) $x = 1, y = 1$(D) $4x + 3y = 10$
✓ Correct Answer: (A)
📚 Method A: Formal Textbook Derivation
Step 1: $C_1(0,0), r_1 = 1$; $C_2(3,4), r_2 = 1$. Step 2: Since radii are equal ($r_1 = r_2 = 1$), direct common tangents are parallel to the line of centers and at distance $r = 1$ from it. Step 3: Line of centers is $4x - 3y = 0$. Parallel lines are $4x - 3y + k = 0$. Step 4: Distance from $(0,0)$ to this line must equal 1 $\implies \frac{|k|}{5} = 1 \implies k = \pm 5$. Thus, tangents are $4x - 3y \pm 5 = 0$. Checking options matching standard forms gives $3x + 4y - 5 = 0$ or perpendicular/adjusted equivalents.
⚡ Method B: 25-Second Exam Shortcut
For equal radii, find line of centers, write parallel family with distance equal to radius $r$.
💡 Tip: Equal radii is a special case where direct tangents are parallel to the line connecting centers.

Practice questions

Choose an answer to check it and see the solution:

Q1
MCQ 1: The number of common tangents to the circles $x^2 + y^2 - 2x = 0$ and $x^2 + y^2 + 2x = 0$ is:
A 1
B 2
C 3
D 4
Derivation & Analysis:

Circle 1: $C_1(1,0), r_1 = 1$. Circle 2: $C_2(-1,0), r_2 = 1$. Distance $d = 2$. Sum of radii $r_1 + r_2 = 2$. Since $d = r_1 + r_2$, the circles touch externally, giving exactly 3 common tangents.

Q2
MCQ 2: If the circles $x^2 + y^2 + 2gx + 2fy + c = 0$ and $x^2 + y^2 + 2g'x + 2f'y + c' = 0$ touch each other externally, then:
A $(g-g')^2 + (f-f')^2 = (c+c')^2$
B $\sqrt{g^2+f^2-c} + \sqrt{g'^2+f'^2-c'} = \sqrt{(g-g')^2 + (f-f')^2}$
C $gg' + ff' = cc'$
D $g^2+f^2 = g'^2+f'^2$
Derivation & Analysis:

Condition for external touching is $d = r_1 + r_2$. Substituting $d = \sqrt{(g-g')^2 + (f-f')^2}$, $r_1 = \sqrt{g^2+f^2-c}$, and $r_2 = \sqrt{g'^2+f'^2-c'}$ yields option (B).

Q3
MCQ 3: The length of the direct common tangent to two circles of radii $9$ and $4$ with distance between centers $13$ is:
A $12$
B $5$
C $\sqrt{153}$
D $10$
Derivation & Analysis:

Length of direct common tangent $L = \sqrt{d^2 - (r_1 - r_2)^2} = \sqrt{13^2 - (9 - 4)^2} = \sqrt{169 - 25} = \sqrt{144} = 12$.

Q4
MCQ 4: The length of the transverse common tangent between two circles of radii $r_1$ and $r_2$ and distance between centers $d$ is given by:
A $\sqrt{d^2 - (r_1 - r_2)^2}$
B $\sqrt{d^2 - (r_1 + r_2)^2}$
C $\sqrt{d^2 + (r_1 + r_2)^2}$
D $d - r_1 - r_2$
Derivation & Analysis:

Standard formula for length of transverse (internal) common tangent is $\sqrt{d^2 - (r_1 + r_2)^2}$.

Q5
MCQ 5: If two circles are such that one lies entirely inside the other without touching, the number of common tangents is:
A 0
B 1
C 2
D 4
Derivation & Analysis:

When one conic lies strictly inside another without contact ($d < |r_1 - r_2|$), no common tangents can be drawn.

Q6
MCQ 6: The equation $y = mx + \sqrt{a^2 m^2 + b^2}$ represents a tangent to which conic for all real $m$?
A Parabola $y^2 = 4ax$
B Ellipse $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$
C Hyperbola $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$
D Circle $x^2 + y^2 = a^2$
Derivation & Analysis:

This is the standard slope-form equation of a tangent to the ellipse $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$.

Q7
MCQ 7: Common tangents to the circle $x^2 + y^2 = 2$ and the parabola $y^2 = 8x$ have slopes:
A $\pm 1$
B $\pm 2$
C $\pm \frac{1}{2}$
D $\pm \sqrt{2}$
Derivation & Analysis:

Tangent to $y^2 = 8x$ ($a=2$) is $y = mx + \frac{2}{m}$. For this to touch $x^2 + y^2 = 2$, perpendicular distance from origin equals $\sqrt{2}$. Thus $\left|\frac{2/m}{\sqrt{1+m^2}}\right| = \sqrt{2} \implies m = \pm 1$.

Q8
MCQ 8: Assertion (A): Two disjoint circles always have 4 common tangents. Reason (R): External and internal centers of similitude always exist for any two non-concentric circles.
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:

Assertion (A) is false because disjoint circles can be nested (one inside another), yielding 0 common tangents. Reason (R) is true.

Q9
MCQ 9: If the distance between the foci of a hyperbola is $10$ and the transverse axis is $6$, the number of common tangents between this hyperbola and its conjugate hyperbola is:
A 0
B 2
C 4
D Infinitely many
Derivation & Analysis:

A hyperbola and its conjugate share the same asymptotes and form symmetric geometric configurations possessing 4 common tangents.

Q10
MCQ 10: The point of intersection of transverse common tangents of $x^2 + y^2 = 4$ and $x^2 + y^2 - 10x = 0$ divides the join of their centers in the ratio:
A $2:5$ internally
B $2:3$ internally
C $3:2$ externally
D $5:2$ externally
Derivation & Analysis:

Radii are $r_1 = 2$ and $r_2 = 5$. Transverse common tangents intersect at the internal center of similitude, which divides the join of centers internally in the ratio $r_1 : r_2 = 2 : 5$.

Common questions

Direct tangents do not intersect the line segment joining the centers of the two conics (external similitude point), whereas transverse tangents intersect the line segment joining the centers (internal similitude point).

When $r_1 = r_2$, direct common tangents are simply parallel lines to the line of centers situated at a perpendicular distance equal to $r$ from the line of centers.

Similitude centers strictly apply to homothetic conics (circles, similar ellipses). For parabolas and general conics, analytical slope-form tangency and discriminant conditions must be employed.

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