Mathematics / Geometry 2D Coordinate Geometry & Conics 100% Free Open Access
Chapter 3 โ€ข Theory & Derivations

The Circle: Tangents, Polars & Systems of Coaxial Circles

Comprehensive mathematical theory of circles: standard, general, and diametric representations; tangency conditions, slope equations, and Joachimsthal's pair of tangents SS_1 = T^2; chord of contact and reciprocal pole and polar theory; radical axis locus S_1 - S_2 = 0, perpendicularity proof, and radical centers; orthogonal circle criteria 2g_1 g_2 + 2f_1 f_2 = c_1 + c_2; and coaxial systems of circles, canonical equations, limiting points, and conjugate orthogonal families.

ยง3.1 Standard, General & Diametric Equations of the Circle

1. The Locus Definition & General Form

A circle is the planar locus of a point $P(x, y)$ that moves such that its Euclidean distance from a fixed center $C(x_0, y_0)$ remains constant and equal to radius $R > 0$: $$(x - x_0)^2 + (y - y_0)^2 = R^2$$ Expanding this central equation yields the general second-degree equation of a circle: $$x^2 + y^2 + 2gx + 2fy + c = 0$$ Completing the squares: $$(x + g)^2 + (y + f)^2 = g^2 + f^2 - c$$ Comparing with the standard form: $$\mathbf{Center:} \; C(-g, -f), \qquad \mathbf{Radius:} \; R = \sqrt{g^2 + f^2 - c}$$ Classification based on $g^2 + f^2 - c$:

  • Real Circle ($g^2 + f^2 > c$): A non-degenerate circle with positive real radius.
  • Point Circle ($g^2 + f^2 = c$): The radius is zero; the locus degenerates to the single point $(-g, -f)$.
  • Imaginary / Virtual Circle ($g^2 + f^2 < c$): No real coordinates satisfy the equation.

2. Circle on a Given Diameter

Let $A(x_1, y_1)$ and $B(x_2, y_2)$ be the endpoints of a diameter. For any point $P(x, y)$ on the circle (other than $A$ or $B$), the inscribed angle $\angle APB = \pi/2$ by Thales' theorem. Hence $\vec{PA} \cdot \vec{PB} = 0$: $$(x - x_1)(x - x_2) + (y - y_1)(y - y_2) = 0$$ This is the celebrated diametric form of a circle!

ยง3.2 Tangents, Normals & Length of Tangents

1. Equation of Tangent and Normal at a Point

For the general circle $S \equiv x^2 + y^2 + 2gx + 2fy + c = 0$, the equation of the tangent at point $P(x_1, y_1)$ on the circle is obtained by the rule of transformation $x^2 \to x x_1, y^2 \to y y_1, 2x \to x + x_1, 2y \to y + y_1$: $$T \equiv x x_1 + y y_1 + g(x + x_1) + f(y + y_1) + c = 0$$ Because the normal passes through the center $(-g, -f)$ and point $(x_1, y_1)$, its equation is: $$\frac{x - x_1}{x_1 + g} = \frac{y - y_1}{y_1 + f} \iff (y_1 + f)(x - x_1) - (x_1 + g)(y - y_1) = 0$$

2. Tangents in Slope Form & Condition of Tangency

For the central circle $x^2 + y^2 = R^2$, a straight line $y = mx + k$ is tangent to the circle if and only if the perpendicular distance from the center $(0, 0)$ to the line equals the radius: $$\frac{|k|}{\sqrt{1 + m^2}} = R \iff k = \pm R \sqrt{1 + m^2}$$ Thus the two parallel tangents with slope $m$ are: $$y = mx \pm R \sqrt{1 + m^2}$$

3. Length of Tangents & Pair of Tangents

The length $L$ of the tangent drawn from an external point $P(x_1, y_1)$ to the circle $S = 0$ is: $$L = \sqrt{S_1} = \sqrt{x_1^2 + y_1^2 + 2gx_1 + 2fy_1 + c}$$ The combined joint equation of the pair of tangents drawn from $P(x_1, y_1)$ to the circle is given by the elegant Joachimsthal relation: $$S S_1 = T^2$$ where $S = x^2 + y^2 + 2gx + 2fy + c$, $S_1 = x_1^2 + y_1^2 + 2gx_1 + 2fy_1 + c$, and $T = xx_1 + yy_1 + g(x+x_1) + f(y+y_1) + c$.

ยง3.3 Chord of Contact, Pole and Polar Theory

1. Chord of Contact

If two tangents are drawn from an external point $P(x_1, y_1)$ touching the circle at $Q$ and $R$, the line segment connecting the points of tangency is the chord of contact. Its equation is identically: $$T \equiv x x_1 + y y_1 + g(x + x_1) + f(y + y_1) + c = 0$$

2. Pole and Polar Theory

Let $P(x_1, y_1)$ be any point (internal, external, or on the circle). If a variable secant line through $P$ intersects the circle at $A$ and $B$, the locus of the intersection of the tangents drawn at $A$ and $B$ is a straight line termed the polar of $P$ with respect to the circle. Point $P$ is called the pole of this line. The equation of the polar of point $P(x_1, y_1)$ is: $$T \equiv x x_1 + y y_1 + g(x + x_1) + f(y + y_1) + c = 0$$

Key Properties of Polars:

  • If $P$ lies outside the circle, the polar is the chord of contact of tangents from $P$.
  • If $P$ lies on the circle, the polar is the tangent line at $P$.
  • If $P$ lies inside the circle, the polar lies entirely outside the circle.
  • Reciprocal Property: If the polar of point $P$ passes through point $Q$, then the polar of point $Q$ passes through point $P$. Such points $P$ and $Q$ are termed conjugate points.

ยง3.4 Radical Axis, Radical Center & Orthogonal Circles

1. The Radical Axis

The radical axis of two non-concentric circles $S_1 \equiv x^2 + y^2 + 2g_1 x + 2f_1 y + c_1 = 0$ and $S_2 \equiv x^2 + y^2 + 2g_2 x + 2f_2 y + c_2 = 0$ is the planar locus of points from which the lengths of tangents drawn to both circles are equal: $$L_1^2 = L_2^2 \iff S_1 - S_2 = 0$$ Subtracting the two equations eliminates the quadratic terms $x^2 + y^2$, yielding a linear equation: $$2(g_1 - g_2)x + 2(f_1 - f_2)y + (c_1 - c_2) = 0$$ Geometric Theorem: The radical axis is always perpendicular to the line joining the centers $C_1(-g_1, -f_1)$ and $C_2(-g_2, -f_2)$ of the two circles!

2. The Radical Center

For three circles $S_1 = 0, S_2 = 0, S_3 = 0$ whose centers are non-collinear, the three radical axes taken in pairs: $$S_1 - S_2 = 0, \qquad S_2 - S_3 = 0, \qquad S_3 - S_1 = 0$$ are concurrent at a unique point termed the radical center. The lengths of tangents drawn from the radical center to all three circles are equal, so it is the center of a circle orthogonal to all three circles!

3. Orthogonal Circles Condition

Two circles intersect orthogonally if their tangents at the points of intersection are perpendicular. By the Pythagorean theorem on the triangle formed by the two centers and a point of intersection: $$C_1 C_2^2 = R_1^2 + R_2^2 \iff (g_1 - g_2)^2 + (f_1 - f_2)^2 = (g_1^2 + f_1^2 - c_1) + (g_2^2 + f_2^2 - c_2)$$ Expanding and simplifying establishes the condition of orthogonality: $$2 g_1 g_2 + 2 f_1 f_2 = c_1 + c_2$$

ยง3.5 Coaxial Systems of Circles & Limiting Points

1. Coaxial Systems of Circles

A system of circles is said to be coaxial if every pair of circles in the system possesses the same common radical axis. If $S_1 = 0$ and $S_2 = 0$ are two members of the system, any circle in the coaxial family is given by: $$S_1 + \lambda S_2 = 0 \quad (\lambda \ne -1), \qquad \text{or} \quad S_1 + \mu(S_1 - S_2) = 0$$ By choosing the common radical axis as the $y$-axis ($x = 0$) and the line of centers as the $x$-axis ($y = 0$), the simplest canonical form of a coaxial system is: $$x^2 + y^2 + 2 k x + c = 0$$ where $c$ is a constant for the entire family and $k$ is a variable parameter identifying individual circles. The center is $(-k, 0)$ and radius is $R = \sqrt{k^2 - c}$.

2. Limiting Points

The limiting points of a coaxial system are the centers of the circles in the system whose radii vanish identically ($R = 0$): $$R^2 = k^2 - c = 0 \iff k = \pm \sqrt{c}$$ Thus, if $c > 0$, there exist two real limiting points: $$L_1(\sqrt{c}, 0), \qquad L_2(-\sqrt{c}, 0)$$ These limiting points are point-circles belonging to the coaxial system. Every circle of the system is orthogonal to any circle passing through the two limiting points!

TIERED UNIVERSITY HONORS PROBLEMS

Step-by-Step Solved Examination Problems

Comprehensive analytical derivations, multi-tier solutions (Foundational, Intermediate Exam, and Honors/Proof Challenge) with complete line-by-line verification.

Tier 1: Foundational Example 3.1: Circle Passing Through Three Points and Tangent at a Given Point

Find the equation of the circle passing through the points $(0, 0)$, $(4, 0)$, and $(0, 6)$. Determine its center, radius, and the equation of the tangent line at the origin $(0, 0)$.

Step 1: Determine Circle Equation
$$x^2 + y^2 + 2gx + 2fy + c = 0 \\ \text{Through } (0, 0): \quad c = 0 \\ \text{Through } (4, 0): \quad 16 + 8g = 0 \implies g = -2 \\ \text{Through } (0, 6): \quad 36 + 12f = 0 \implies f = -3 \\ \text{Circle: } x^2 + y^2 - 4x - 6y = 0$$

Substituting the three points determines the unique values of $g, f, c$.

Step 2: Find Center and Radius
$$\text{Center: } (-g, -f) = (2, 3) \\ \text{Radius: } R = \sqrt{g^2 + f^2 - c} = \sqrt{(-2)^2 + (-3)^2 - 0} = \sqrt{4 + 9} = \sqrt{13}$$

The circle is centered at $(2, 3)$ with radius $\sqrt{13}$.

Step 3: Tangent Line at the Origin $(0, 0)$
$$T \equiv x(0) + y(0) - 2(x + 0) - 3(y + 0) = 0 \implies -2x - 3y = 0 \implies 2x + 3y = 0$$

Using the transformation rule $T = 0$ gives the tangent line $2x + 3y = 0$.

Final Answer & Physical Insight

\text{Circle: } x^2 + y^2 - 4x - 6y = 0; \quad \text{Center: } (2, 3); \quad R = \sqrt{13}; \quad \text{Tangent at } (0, 0): 2x + 3y = 0

Tier 2: Intermediate Exam Example 3.2: Radical Axis and Condition of Orthogonality

Given two circles $S_1 \equiv x^2 + y^2 - 6x - 4y + 9 = 0$ and $S_2 \equiv x^2 + y^2 + 4x + 6y - 7 = 0$: (a) Find the equation of their radical axis. (b) Verify that the radical axis is perpendicular to the line of centers. (c) Find the value of $\lambda$ such that the circle $x^2 + y^2 + \lambda x - 2y + 5 = 0$ is orthogonal to $S_1$.

Step 1: Compute Radical Axis $S_1 - S_2 = 0$
$$(x^2 + y^2 - 6x - 4y + 9) - (x^2 + y^2 + 4x + 6y - 7) = 0 \\ -10x - 10y + 16 = 0 \implies 5x + 5y - 8 = 0$$

The radical axis is $5x + 5y - 8 = 0$, having slope $m_{ ext{rad}} = -5/5 = -1$.

Step 2: Verify Perpendicularity to Line of Centers
$$C_1 = (3, 2), \quad C_2 = (-2, -3) \\ m_{\text{centers}} = \frac{-3 - 2}{-2 - 3} = \frac{-5}{-5} = 1 \\ m_{\text{rad}} \times m_{\text{centers}} = (-1)(1) = -1 \implies \text{Strictly Perpendicular!}$$

The product of slopes is $-1$, verifying perpendicularity.

Step 3: Condition of Orthogonality for $S_1$ and $S_3$
$$S_1: g_1 = -3, f_1 = -2, c_1 = 9 \\ S_3: g_3 = \lambda/2, f_3 = -1, c_3 = 5 \\ 2 g_1 g_3 + 2 f_1 f_3 = c_1 + c_3 \implies 2(-3)\left(\frac{\lambda}{2}\right) + 2(-2)(-1) = 9 + 5 \\ -3\lambda + 4 = 14 \implies -3\lambda = 10 \implies \lambda = -\frac{10}{3}$$

Applying $2g_1 g_2 + 2f_1 f_2 = c_1 + c_2$ yields $\lambda = -10/3$.

Final Answer & Physical Insight

\text{Radical Axis: } 5x + 5y - 8 = 0; \quad m_1 m_2 = -1 \implies \text{Perpendicular}; \quad \lambda = -\frac{10}{3}

Tier 3: Honors / Proof Challenge Example 3.3: Coaxial Family Limiting Points and Orthogonal Conjugate System

A coaxial system of circles is determined by $S_1 \equiv x^2 + y^2 - 4x - 6y + 9 = 0$ and the radical axis $x - y + 1 = 0$. (a) Write the general equation of the coaxial family. (b) Find the exact coordinates of the limiting points of the system. (c) Derive the equation of the orthogonal conjugate coaxial system passing through the limiting points.

Step 1: Formulate the Coaxial System $S + \lambda L = 0$
$$x^2 + y^2 - 4x - 6y + 9 + \lambda(x - y + 1) = 0 \\ x^2 + y^2 + (\lambda - 4)x - (\lambda + 6)y + (\lambda + 9) = 0$$

Here $g = \frac{\lambda - 4}{2}$, $f = -\frac{\lambda + 6}{2}$, $c = \lambda + 9$.

Step 2: Find Limiting Points via $R^2 = g^2 + f^2 - c = 0$
$$\left(\frac{\lambda - 4}{2}\right)^2 + \left(\frac{\lambda + 6}{2}\right)^2 - (\lambda + 9) = 0 \\ \frac{\lambda^2 - 8\lambda + 16 + \lambda^2 + 12\lambda + 36}{4} - (\lambda + 9) = 0 \\ \frac{2\lambda^2 + 4\lambda + 52}{4} - (\lambda + 9) = 0 \implies \frac{\lambda^2 + 2\lambda + 26}{2} - (\lambda + 9) = 0 \\ \lambda^2 + 2\lambda + 26 - 2\lambda - 18 = 0 \implies \lambda^2 + 8 = 0$$

Notice $\lambda^2 = -8$, meaning the limiting points are imaginary! When limiting points are imaginary, the coaxial circles intersect in real points $A, B$.

Step 3: Find Real Common Points of Intersection
$$\text{Substitute } y = x + 1 \text{ into } S_1: \quad x^2 + (x+1)^2 - 4x - 6(x+1) + 9 = 0 \\ x^2 + x^2 + 2x + 1 - 4x - 6x - 6 + 9 = 0 \implies 2x^2 - 8x + 4 = 0 \implies x^2 - 4x + 2 = 0 \\ x = 2 \pm \sqrt{2}, \quad y = 3 \pm \sqrt{2}$$

The common real intersection points of the intersecting coaxial family are $P_1(2 + \sqrt{2}, 3 + \sqrt{2})$ and $P_2(2 - \sqrt{2}, 3 - \sqrt{2})$. The orthogonal conjugate system has these two points as its real limiting points!

Final Answer & Physical Insight

\text{Coaxial System: } x^2 + y^2 + (\lambda - 4)x - (\lambda + 6)y + (\lambda + 9) = 0; \quad \text{Limiting Points: Imaginary (Intersecting family)}; \quad \text{Common Points: } (2 \pm \sqrt{2}, 3 \pm \sqrt{2})