Interactive Lesson: Finding Transverse Common Tangents An interactive, step-by-step guide to solving a common geometry problem

Interactive Lesson: Finding Transverse Common Tangents

An interactive, step-by-step guide to solving a common geometry problem.

The Problem

Our goal is to find the equations of the transverse common tangents of the two circles given by the following equations:

Circle 1: \(x^2 + y^2 + 4x - 6y + 4 = 0\)

Circle 2: \(x^2 + y^2 - 4x - 10y + 28 = 0\)

The visualization on the left shows these two circles plotted on a graph. Use the 'Next' button to walk through the solution step-by-step. Each step in the solution will correspond to a change in the visualization.

Step 1: Find Centers and Radii

First, we analyze the general equation of each circle, \(x^2 + y^2 + 2gx + 2fy + c = 0\), to find its center \((-g, -f)\) and radius \(r = \sqrt{g^2 + f^2 - c}\).

Circle 1

\(x^2 + y^2 + 4x - 6y + 4 = 0\)

Center \(O_1 = (-2, 3)\)

Radius \(r_1 = 3\)

Circle 2

\(x^2 + y^2 - 4x - 10y + 28 = 0\)

Center \(O_2 = (2, 5)\)

Radius \(r_2 = 1\)

On the chart, the centers of the circles, \(O_1\) and \(O_2\), are now marked.

Step 2: Find the Tangent Intersection Point (P)

For transverse tangents, the intersection point \(P\) divides the line segment connecting the centers \(O_1O_2\) internally in the ratio of the radii, \(r_1:r_2\), which is \(3:1\).

Using the internal division formula:

\(P(x,y) = \left(\frac{n x_1 + m x_2}{m+n}, \frac{n y_1 + m y_2}{m+n}\right)\)

We get \(P = \left(1, \frac{9}{2}\right)\).

The chart now shows the line segment connecting the centers and the crucial intersection point \(P\).

Step 3: General Equation of the Tangent

Any line passing through \(P(1, 9/2)\) can be written using the point-slope form \(y - y_p = m(x - x_p)\).

\(y - \frac{9}{2} = m(x - 1)\)

Rearranging this into the general form \(Ax + By + C = 0\), we get:

\(2mx - 2y + (9 - 2m) = 0\)

Our next task is to find the specific value(s) of the slope \(m\).

Step 4: Use the Distance Formula

The fundamental property of a tangent is that the distance from the center of a circle to the tangent line is equal to the circle's radius.

We'll set the distance from \(O_1(-2, 3)\) to our general tangent line equal to \(r_1 = 3\).

\(d = \frac{|Ax_0 + By_0 + C|}{\sqrt{A^2 + B^2}}\)

Substituting our values leads to:

\(3 = \frac{|2m(-2) - 2(3) + (9 - 2m)|}{\sqrt{(2m)^2 + (-2)^2}}\)

After simplifying, we get the equation:

\(2\sqrt{m^2 + 1} = |1 - 2m|\)

Step 5: Solve for the Slope(s)

To solve for \(m\), we square both sides of the equation from the previous step:

\(4(m^2 + 1) = (1 - 2m)^2\)

\(4m^2 + 4 = 1 - 4m + 4m^2\)

This simplifies to \(4 = 1 - 4m\), which gives us \(m = -\frac{3}{4}\).

A Special Case: The Vertical Tangent

Our algebraic method only produced one slope. This often indicates the existence of a vertical tangent (with an undefined slope), which the \(y=mx+c\) form cannot represent. By checking if a vertical line \(x=1\) through point \(P\) is tangent to both circles, we find that it is!

Step 6: The Final Equations

We now have everything needed to state the equations of the two transverse common tangents.

Tangent 1 (with \(m = -3/4\))

\(3x + 4y - 21 = 0\)

Tangent 2 (the vertical line)

\(x = 1\)

The final tangent lines have been drawn on the chart, completing the solution visually.

Finding Transverse Common Tangents of Two Circles IPE 2A Exercise 1(e) II 4

📐 Finding Transverse Common Tangents of Two Circles

🎯 Problem:

Find the transverse common tangents of the circles:

Circle 1: x² + y² - 4x - 10y + 28 = 0
Circle 2: x² + y² + 4x - 6y + 4 = 0

Convert to Standard Form

Rewrite both equations in the form (x - h)² + (y - k)² = r²

For Circle 1: x² + y² - 4x - 10y + 28 = 0

Group x and y terms: (x² - 4x) + (y² - 10y) = -28

Complete the square:

For x terms: x² - 4x + 4 = (x - 2)²

For y terms: y² - 10y + 25 = (y - 5)²

(x - 2)² + (y - 5)² = -28 + 4 + 25 = 1

Circle 1: (x - 2)² + (y - 5)² = 1

Center: C₁(2, 5), Radius: r₁ = 1

For Circle 2: x² + y² + 4x - 6y + 4 = 0

Group x and y terms: (x² + 4x) + (y² - 6y) = -4

Complete the square:

For x terms: x² + 4x + 4 = (x + 2)²

For y terms: y² - 6y + 9 = (y - 3)²

(x + 2)² + (y - 3)² = -4 + 4 + 9 = 9

Circle 2: (x + 2)² + (y - 3)² = 9

Center: C₂(-2, 3), Radius: r₂ = 3

Analyze Circle Configuration

Calculate the distance between centers:

d = √[(x₂ - x₁)² + (y₂ - y₁)²]

d = √[(2 - (-2))² + (5 - 3)²]

d = √[4² + 2²] = √[16 + 4] = √20 = 2√5 ≈ 4.47

Circle Relationship Analysis:

Condition	Relationship	Common Tangents
d > r₁ + r₂	Separate circles	4 tangents (2 transverse, 2 direct)
d = r₁ + r₂	Externally tangent	3 tangents
|r₁ - r₂| < d < r₁ + r₂	Intersecting	2 tangents

Our case: d = 2√5 ≈ 4.47 and r₁ + r₂ = 1 + 3 = 4

Since d > r₁ + r₂, the circles are separate.

Therefore, there are 4 common tangents (2 transverse + 2 direct).

Set Up Tangent Line Equation

Any line can be written as: ax + by + c = 0

For convenience, we normalize so that a² + b² = 1

Distance from Point to Line Formula:

Distance from point (x₀, y₀) to line ax + by + c = 0 is:

d = |ax₀ + by₀ + c| / √(a² + b²)

Since we normalized with a² + b² = 1:

d = |ax₀ + by₀ + c|

Apply Tangent Conditions

For a line to be tangent to a circle, the distance from the center to the line must equal the radius.

🔍 Key Insight: Transverse vs Direct Tangents

Transverse tangents: The circles are on opposite sides of the tangent line.

Direct tangents: The circles are on the same side of the tangent line.

For transverse tangents, the signs of the distances are opposite:

Distance from C₁(2, 5) to line = r₁ = 1: 2a + 5b + c = 1 ... (1)

Distance from C₂(-2, 3) to line = r₂ = 3: -2a + 3b + c = -3 ... (2)

Notice the opposite signs (1 and -3) indicating the circles are on opposite sides of the tangent.

Solve the System of Equations

We have three equations with three unknowns:

2a + 5b + c = 1 ... (1)

-2a + 3b + c = -3 ... (2)

a² + b² = 1 ... (3)

Subtract equation (2) from equation (1):

(2a + 5b + c) - (-2a + 3b + c) = 1 - (-3)

2a + 5b + c + 2a - 3b - c = 4

4a + 2b = 4

2a + b = 2

b = 2 - 2a ... (4)

Substitute (4) into equation (1):

2a + 5(2 - 2a) + c = 1

2a + 10 - 10a + c = 1

-8a + c = -9

c = 8a - 9 ... (5)

Use Normalization Condition

Substitute equations (4) and (5) into the normalization condition a² + b² = 1:

a² + (2 - 2a)² = 1

a² + (4 - 8a + 4a²) = 1

a² + 4 - 8a + 4a² = 1

5a² - 8a + 3 = 0

Solve the quadratic equation:

Using the quadratic formula or factoring:

5a² - 8a + 3 = (5a - 3)(a - 1) = 0

a = 3/5 or a = 1

Find Both Tangent Lines

Case 1: a = 3/5

From equation (4): b = 2 - 2(3/5) = 2 - 6/5 = 4/5

From equation (5): c = 8(3/5) - 9 = 24/5 - 45/5 = -21/5

Tangent Line 1: (3/5)x + (4/5)y + (-21/5) = 0

Multiplying by 5: 3x + 4y - 21 = 0

Case 2: a = 1

From equation (4): b = 2 - 2(1) = 0

From equation (5): c = 8(1) - 9 = -1

Tangent Line 2: x + 0y + (-1) = 0

Simplifying: x - 1 = 0 or x = 1

Verify the Solution

Let's verify that our tangent lines are correct:

For Tangent 1: 3x + 4y - 21 = 0

Distance from C₁(2, 5): |3(2) + 4(5) - 21| / √(3² + 4²) = |6 + 20 - 21| / 5 = 1 / 5 = 1/5

Wait, this should equal r₁ = 1. Let me recalculate...

Actually: |3(2) + 4(5) - 21| / √(9 + 16) = |26 - 21| / 5 = 5/5 = 1 ✓

Distance from C₂(-2, 3): |3(-2) + 4(3) - 21| / 5 = |-6 + 12 - 21| / 5 = 15/5 = 3 ✓

For Tangent 2: x = 1

Distance from C₁(2, 5): |2 - 1| = 1 ✓

Distance from C₂(-2, 3): |-2 - 1| = 3 ✓

✅ Both tangent lines satisfy the distance conditions!

🎉 Final Answer

The transverse common tangents of the two circles are:

Tangent 1: 3x + 4y - 21 = 0

Tangent 2: x = 1

📚 Teaching Summary

Key concepts students should understand:

• Converting general circle equations to standard form
• Understanding the geometric meaning of transverse tangents
• Setting up distance conditions for tangency
• Solving systems of equations with constraints
• Verifying solutions geometrically

Transverse Common Tangents of Two Circles PE 2A Exercise 1(e) II 4

🔵 Transverse Common Tangents of Two Circles

Circle 1: (x-2)² + (y-5)² = 1

Circle 2: (x+2)² + (y-3)² = 9

Transverse Common Tangents

🎯 What are Transverse Common Tangents?

Transverse common tangents are lines that:

• Touch both circles at exactly one point each
• Cross between the circles (circles are on opposite sides)
• For two separate circles, there are exactly 2 transverse tangents

Click the buttons above to see the step-by-step construction!

📐 Mathematical Solution:

Circle 1: x² + y² - 4x - 10y + 28 = 0 → (x-2)² + (y-5)² = 1

Circle 2: x² + y² + 4x - 6y + 4 = 0 → (x+2)² + (y-3)² = 9

Transverse Common Tangents:

• Tangent 1: 3x + 4y - 21 = 0
• Tangent 2: x = 1