Two circles are said to be **orthogonal** if they intersect at a right angle (90°). This means that at their point of intersection, the tangents to the two circles are perpendicular.
Let the two circles be:
S₁ ≡ x² + y² + 2g₁x + 2f₁y + c₁ = 0
S₂ ≡ x² + y² + 2g₂x + 2f₂y + c₂ = 0
Their centers are C₁(-g₁, -f₁) and C₂(-g₂, -f₂), and their radii are r₁ = sqrt(g₁² + f₁² - c₁) and r₂ = sqrt(g₂² + f₂² - c₂).
If the circles intersect orthogonally, the triangle formed by the two centers and an intersection point P is a right-angled triangle. By the Pythagorean theorem, (C₁C₂)² = r₁² + r₂².
This leads to the **condition for orthogonality**:
2g₁g₂ + 2f₁f₂ = c₁ + c₂
g₁, f₁, c₁ and g₂, f₂, c₂ and plug them into the condition.
The **radical axis** of two circles is the locus of a point from which the lengths of the tangents drawn to the two circles are equal.
Let the two circles be S₁ = 0 and S₂ = 0 (where coefficients of x² and y² are 1).
The equation of the radical axis is:
S₁ - S₂ = 0
Example:
S₁ ≡ x² + y² + 4x + 6 = 0
S₂ ≡ x² + y² - 2x + 1 = 0
Radical Axis: S₁ - S₂ = (x² + y² + 4x + 6) - (x² + y² - 2x + 1) = 0
4x + 6 - (-2x + 1) = 0
6x + 5 = 0 (This is a straight line).
Properties of the Radical Axis:
The **radical centre** of three circles is the single point from which the lengths of tangents drawn to all three circles are equal.
It is found by taking the three circles in pairs and finding their radical axes. The point of intersection of these three radical axes is the radical centre.
Given circles S₁ = 0, S₂ = 0, and S₃ = 0:
L₁₂ ≡ S₁ - S₂ = 0L₂₃ ≡ S₂ - S₃ = 0L₁₂ = 0 and L₂₃ = 0 simultaneously.The resulting point (x, y) is the radical centre. (The third radical axis L₁₃ ≡ S₁ - S₃ = 0 will automatically pass through this point).
Let S₁ = 0 and S₂ = 0 be two intersecting circles.
The equation of *any* circle passing through the intersection points of S₁ and S₂ is given by:
S₁ + kS₂ = 0 (where k is a parameter and k ≠ -1)
If k = -1, the equation becomes S₁ - S₂ = 0, which is the equation of the radical axis (their common chord).
Let S = 0 be a circle and L = 0 be a straight line.
The equation of *any* circle passing through the intersection points of S and L is given by:
S + kL = 0 (where k is a parameter)
This is a crucial topic. We want to find the condition for a straight line y = mx + c to just touch (be tangent to) a given conic section.
The method is to substitute y = mx + c into the conic's equation. This will result in a quadratic equation in x. For the line to be a tangent, the two intersection points must be coincident, meaning the quadratic equation must have equal roots.
The condition for equal roots of Ax² + Bx + C = 0 is B² - 4AC = 0.
Applying this method gives the following standard conditions:
Condition: c² = a²(1 + m²)
Equation of tangent:y = mx ± a * sqrt(1 + m²)
Condition: c = a/m
Equation of tangent:y = mx + a/m
Condition: c² = a²m² + b²
Equation of tangent:y = mx ± sqrt(a²m² + b²)
Condition: c² = a²m² - b²
Equation of tangent:y = mx ± sqrt(a²m² - b²)
MEMORIZE THIS TABLE. It is the most important part of this unit.
| Conic | Equation | Line | Condition of Tangency | Equation of Tangent |
|---|---|---|---|---|
| Circle | x² + y² = a² |
y = mx + c |
c² = a²(1 + m²) |
y = mx ± a*sqrt(1+m²) |
| Parabola | y² = 4ax |
y = mx + c |
c = a/m |
y = mx + a/m |
| Ellipse | x²/a² + y²/b² = 1 |
y = mx + c |
c² = a²m² + b² |
y = mx ± sqrt(a²m² + b²) |
| Hyperbola | x²/a² - y²/b² = 1 |
y = mx + c |
c² = a²m² - b² |
y = mx ± sqrt(a²m² - b²) |
2g₁g₂ + 2f₁f₂ = c₁ + c₂S₁ - S₂ = 0S + kL = 0