Midpoint and Perpendicular Bisector

• The midpoint of a line segment formula: Midpoint = ((x1+x2)/2, (y1+y2)/2), where (x1, y1) and (x2, y2) are the end points of the line segment.
• A perpendicular bisector of a line segment is a line that is perpendicular to the line segment and passes through the midpoint of the line segment.

Example 1

• Line segment PQ is the diameter of a circle
• P is at (2,3), Q is at (6,7)
• The midpoint of the line is the center of the circle
• Midpoint of PQ: ((2+6)/2, (3+7)/2) = (4,5)
• The center of the circle is (4,5)

Example 2

• Line l₁ passes through the center of the circle (x1, y1)
• l₁ is perpendicular to line segment AB
• Line segment AB is the diameter of the circle
• The midpoint of the line is the center of the circle
• Midpoint of AB: ((3+7)/2, (6+4)/2) = (5,5)
• The center of the circle is C (5,5)
• Given line l₁ is perpendicular to line segment, then the gradient of l₁ is m₂ = -1/m1
• Gradient of the line segment AB: m₁ = (6-4)/(7-3) = 1/2
• The gradient of the line l₁ is m₂ = -2
• The equation of the line l₁ is y = -2x + 15

Equation of a Circle

• The equation of any given circle with center (a, b) and radius r: (x-a)² + (y-b)² = r²
• All points on the circumference have the same distance to its center, the radius of the equation of the circle is derived using Pythagoras’ theorem.

Equation Forms

• Circles may have equations in the form of ax² + by² + cx + dy + e = 0
• Equations in this form are simplified versions of (x − a)² + (y − b)² = r²
• To convert from ax² + by² + cx + dy + e = 0 to (x − a)² + (y − b)² = r², complete the square.

Example 3

• Need to find the center and radius of a circle with equation x² + y² - 4x - 6y - 3 = 0
• Rearrange the equation: x² - 4x + y² - 6y = 3
• Need to complete the square of x² - 4x and y² - 6y
• x² - 4x = (x - 2)² - 4
• y² - 6y = (y - 3)² - 9
• Substitute into the equation of the circle: (x - 2)² - 4 + (y - 3)² - 9 = 3, which simplifies to (x - 2)² + (y - 3)² = 16.
• The center of the circle is at (2,3) and the radius is r = √16 = 4

Intersections of Straight Lines and Circles

• Use methods to find intersection points of a straight line and a circle similar to finding intersections between a line and another curve.
• There can be zero, one, or two intersection points.

Example 4

• Determine the x-coordinates of the intersection points of the line l₁: y = 2x + 3 and the circle (x - 3)² + (y - 4)² = 9
• Solve the equations simultaneously
• (x - 3)² + (y - 4)² = 9 and y = 2x + 3
• Substituting, get (x - 3)² + (2x + 3 - 4)² = 9
• Leads to (x - 3)² + (2x - 1)² = 9
• Simplified equation: 5x² - 10x + 1 = 0
• Use the quadratic formula to solve for x, resulting in x = (5 + 2√5) / 5 and x = (5 - 2√5) / 5

Example 5

• The task is to find the equation of the tangent line l₁ to the circle at point P
• The circle has the equation x² + y² - 10x - 10y + 45 = 0

Methodology

• Complete the square to find the center of the circle
• (x - 5)² - 25 + (y - 5)² - 25 + 45 = 0
• Simplified form: (x - 5)² + (y - 5)² = 5
• The center of the circle is at C(5,5)
• Calculate the gradient of CP, m1 = (5-4)/(5-3) = 1/2
• The tangent is perpendicular to the radius at the intersection point
• Line l₁ is perpendicular to CP
• The gradient of l₁ is m2 = -2
• Gradient of l1, m2 = -2, and line passes through P(3,4)
• Equation of line l₁ y − 4 = −2(x − 3), which simplifies to y = −2x + 10

Circles and Triangles

• A circle that passes through the vertices of a triangle is its circumcircle
• The perpendicular bisectors of each side of the triangle intersect at the center of the circle, known as the circumcenter.
• In a right-angled triangle, the hypotenuse is the diameter of the circumcircle.
• The angle in a semi-circle is always 90°.

Tangent and Chord Properties

• The tangent to the circle is perpendicular to the circle at the point of intersection
• The perpendicular bisector to a chord of the circle will pass through the circle

Learn about the midpoint formula to find the center of a line segment. Discover how a perpendicular bisector intersects a line at its midpoint, forming a right angle. Explore examples using line segments as diameters of circles.