Circles - Tangent Lines and Theorems

Tangent Line Properties

• Tangent line is always perpendicular to the radius at the point of tangency.
• This forms a right triangle.
• Missing lengths can be found using the Pythagorean theorem.

Theorem Application

• Theorem works both ways:
  • If line AB is tangent to the circle at point P, then it's perpendicular to the radius.
  • If a line is perpendicular to the radius, then it's a tangent.

Angle Relationships

• Two angles formed by the tangent and radii at the point of tangency are congruent.
• These angles measure 90 degrees.
• Given three angles in a quadrilateral, the missing angle can be found by setting the sum equal to 360 degrees.
  • Example: $x + 90 + 135 + 90 = 360$

Example: Checking for Tangency

• To check if a line (MP) is tangent to a circle with center N, verify if the angle NMP is a right angle.
• Check if the sum of angles in triangle NMP equals 180 degrees.
  • $58 + 33 + \angle NMP = 180$
• If the calculated angle NMP is not 90 degrees, then line MP is not tangent.
• Example Calculation:
  • $180 - 58 - 33 = 89$
  • Since $89 \neq 90$, line MP is not tangent.

Example: Finding Lengths with Tangency Assumption

• If segment LK is tangent to circle N at point L, then angle NLK is a right angle.
• This implies dealing with a right triangle, allowing the use of the Pythagorean theorem.
• Using the Pythagorean theorem: $c^2 = a^2 + b^2$
  • $c$ is the hypotenuse (opposite the right angle).
  • $a$ and $b$ are the other two sides.
• Identify the lengths:
  • $a = 9$ (radius)
  • $b = x + 5$
  • $c = 9 + x - 1 = x + 8$
• Setup the equation:
  • $(x+8)^2 = 9^2 + (x+5)^2$
• Expand and simplify:
  • $x^2 + 16x + 64 = 81 + x^2 + 10x + 25$
• Combine like terms:
  • $x^2 + 16x + 64 = x^2 + 10x + 106$
  • $6x = 42$
  • $x = 7$
• Remember to answer the original question. If they ask for KN (where KN = x - 1 + 9), then:
  • $KN = 7 - 1 + 9 = 15$

Useful Formula

• Memorize the formula for squaring a binomial: $(a + b)^2 = a^2 + 2ab + b^2$
• It's consistent regardless of the variables used.

Tangent Point Example

• Given a line equation $3x - 4y = 19$ tangent to a circle centered at $(2, 3)$, find the tangent point.
• Convert the line equation to slope-intercept form $(y = mx + b)$.
  • $-4y = -3x + 19$
  • $y = \frac{3}{4}x - \frac{19}{4}$
• Find the slope of the line perpendicular to it (the radius):
  • The slope will be the opposite reciprocal: $m_{\perp} = -\frac{4}{3}$
• Use the point-slope form to find the equation of the line passing through the center $(2, 3)$ with the perpendicular slope.
• $y - y<em>1 = m(x - x</em>1)$
• $y - 3 = -\frac{4}{3}(x - 2)$
• $y = -\frac{4}{3}x + \frac{8}{3} + 3$
• $y = -\frac{4}{3}x + \frac{17}{3}$
• Convert this to standard form: $4x + 3y = 17$

Finding the Tangent Point

• Solve the system of equations formed by the tangent line and the line containing the radius to find the intersection point (tangent point).
• Use elimination method:
  • $3x - 4y = 19$
  • $4x + 3y = 17$
• Multiply equations to eliminate a variable:
  • Multiply the first equation by 3 and the second by 4:
    • $9x - 12y = 57$
    • $16x + 12y = 68$
• Add the equations to eliminate $y$:
  • $25x = 125$
  • $x = 5$
• Substitute $x = 5$ back into either equation to find $y$:
  • Using $4x + 3y = 17$:
    • $4(5) + 3y = 17$
    • $20 + 3y = 17$
    • $3y = -3$
    • $y = -1$
• The tangent point is $(5, -1)$.

Finding a Second Tangent Line

• Given a circle and a tangent line, find the equation of a second tangent line with the same slope.
• The two tangent lines are parallel.
• Use a point on the second tangent line and the slope.
• If the second point of tangency is at $(−1,7)$, then using slope $3/4$:
• $y - 7 = \frac{3}{4}(x + 1)$
• $y = \frac{3}{4}x + \frac{3}{4} + 7$
• $y = \frac{3}{4}x + \frac{31}{4}$

Congruent Triangles and Tangents

• Given two tangent lines from the same external point to a circle, the triangles formed by radii to the tangent points are congruent.
• These tangent lines are perpendicular to the radii, forming right angles.
• Hypotenuse Leg (HL) Theorem: If the hypotenuse and a leg of one right triangle are congruent to the corresponding parts of another right triangle, the triangles are congruent.
• Corresponding Parts of Congruent Triangles are Congruent (CPCTC):
  • If \[\triangle OYZ \cong \triangle OXZ], then segments YZ = XZ.
  • Angle bisector, e.g. zt is an angle bisector because \[\angle TXZ = \angle TYZ]

Example: Using Congruence

• If TX = 12 and TZ = 20, use triangles to derive length of XZ/ZY with Pythagorean theorem.
• $\sqrt{20^2 - 12^2} = 16$
• $XZ = ZY = 16$

Two Tangent Theorem

• If two segments with a common endpoint exterior to the circle are tangent to the circle, then the segments are congruent.
• If AB and AC are tangent to the circle from point A, then AB = AC.
• Tangent segments from the same external point are always congruent.

Satellite Communication Application

• Satellite requires line of sight for communication with ground stations; ground stations are points of tangency.
• Earth's radius: 6,371 km.
• Satellite altitude: 35,786 km above Earth.
• Problem: Find the amount of time needed for a signal to travel from one station to the satellite and then to the other station.

Solution Steps

• Use the Pythagorean Theorem to find this distance.
• Total Length from the earth's center to the satellite is \[6371 + 35786 = 42157km \]
• Segment from the ground station = \[ (42157)^2 - (6371)^2 = 41,673 km\]
• Once we calculate the length, we need to calculate the amount of time needed for a signal to go from one station up to the satellite and down to the other station.
• Use the formula: $distance = speed \times time$
• Signal speed: 300,000 km/second.
• $41673 = 300000 \times t$
• $t \approx 0.14 seconds$

Perimeter Example

• Given a quadrilateral ABCD circumscribed around a circle, find its perimeter.
• Utilize that tangent segments from the same point are congruent. From that, we can determine values like AD, BC, etc.
• Add all side lengths to find the perimeter.