[Circle Equations: Practice Lab ③] Relative Positions of Two Circles, Common Chords & Circumference Bisection: 15 Killer Problems & Self-Diagnosis
In our previous essay, [Circle Concept Masterclass ③], we broke free from algebraic brute-force systems: "Never solve for quadratic intersection coordinates directly; express pencil families through the identity $C_1 + kC_2 = 0$ and eliminate quadratic terms via $k = -1$ to extract the Radical Axis (Common Chord) in seconds." We also mastered the perpendicular bisector symmetry of the line of centers and the center-passing condition for circumference bisection.
However, competitive exams and standardized assessments (such as IB Math AA HL, A-Level Pure Mathematics, and AMC 10/12) test geometric rigor at a deeper level. They demand that you evaluate minimal area circles constructed on common chords, analyze concentric loci of chord midpoints under parameter rotations, locate the Radical Center where three common chords concur, and solve orthogonal circle intersections and circumference bisection extrema.
In this final Circle Practice Lab, we dissect 15 Advanced Killer Problems and their 1:1 Parallel Diagnosis Variants. Sketch the auxiliary line of centers and right triangles on your own scratchpad first, then open each toggle to compare your reasoning with Yul's Pro-Tips.
1. Radical Axis Is Strictly $C_1 - C_2 = 0$: Subtracting general forms instantly eliminates $x^2$ and $y^2$, delivering the unique linear equation through both intersection points.
2. Line of Centers Is the Perpendicular Bisector: $\overline{O_1 O_2} \perp \overline{AB}$, meaning common chord length reduces immediately to Pythagoras: $\text{Chord} = 2\sqrt{r_1^2 - d^2}$.
3. Circumference Bisection Geometry: For circle $C_1$ to bisect the circumference of circle $C_2$, the common chord must act as a diameter of $C_2$—passing directly through the center of $C_2$.
Theme 1. Radical Axis & Chord Length Pythagorean Triangles (01 ~ 04)
[Problem 01] Determining Parameter Value for a Fixed Common Chord Length
The common chord of circles $C_1: x^2 + y^2 = 20$ and $C_2: (x - 2a)^2 + (y - a)^2 = 5a^2 - 10a + 25$ has a length of $4\sqrt{2}$. Find the value of the positive constant $a$.
π Solution & Step-by-Step Breakdown
Answer: $a = 2$
[Solution]
1) Write both circles in general form:
$C_1: x^2 + y^2 - 20 = 0$
$C_2: x^2 + y^2 - 4ax - 2ay + 10a - 25 = 0$
2) Subtract to find the Radical Axis ($C_1 - C_2 = 0$):
$$4ax + 2ay - 10a + 5 = 0$$
3) Because the chord length is $4\sqrt{2}$, half the chord length is $2\sqrt{2}$.
For $C_1$, center is $(0, 0)$ and radius is $r = \sqrt{20}$. By Pythagoras:
$$d^2 + (2\sqrt{2})^2 = r^2 \implies d^2 + 8 = 20 \implies d^2 = 12 \implies d = 2\sqrt{3}$$
4) Compute distance $d$ from $(0, 0)$ to the Radical Axis:
$$d = \dfrac{|-10a + 5|}{\sqrt{(4a)^2 + (2a)^2}} = \dfrac{5|2a - 1|}{2\sqrt{5}a} = 2\sqrt{3}$$
Squaring and solving for positive $a$ yields $\mathbf{a = 2}$.
Do not find the coordinates of intersection points. Always subtract to get the linear Radical Axis $C_1 - C_2 = 0$, then set up the Pythagorean relation $d = \sqrt{r^2 - (\text{chord}/2)^2}$ using the simpler circle centered at the origin.
π― [Self-Diagnosis Variant 01] (Equal Radii Chord Length)
Circles $x^2 + y^2 = 10$ and $(x - k)^2 + (y - k)^2 = 10$ share a chord of length $2\sqrt{6}$. Find the positive constant $k$.
View Variant Solution
Breakdown: Half chord $= \sqrt{6}, r = \sqrt{10} \implies d = \sqrt{10 - 6} = 2$. With equal radii, distance between centers is $2d = 4$. Center distance: $\sqrt{2k^2} = 4 \implies k = \mathbf{2\sqrt{2}}$.
[Problem 02] Minimum Area Circle Passing Through Two Intersection Points
Find the circumference of the circle having minimum area among all circles passing through the intersection points of $C_1: x^2 + y^2 - 4x + 2y - 4 = 0$ and $C_2: x^2 + y^2 + 2x - 6y + 1 = 0$.
π Solution & Step-by-Step Breakdown
Answer: $\sqrt{11}\pi$
[Solution]
1) The circle passing through two intersection points that achieves minimal area is the one whose diameter is precisely the common chord.
2) Radical Axis ($C_1 - C_2 = 0$):
$$(x^2 + y^2 - 4x + 2y - 4) - (x^2 + y^2 + 2x - 6y + 1) = 0 \implies 6x - 8y + 5 = 0$$
3) For $C_1$, center is $(2, -1)$ and radius is $r_1 = \sqrt{2^2 + (-1)^2 - (-4)} = 3$.
Distance from $(2, -1)$ to $6x - 8y + 5 = 0$:
$$d = \dfrac{|6(2) - 8(-1) + 5|}{\sqrt{6^2 + (-8)^2}} = \dfrac{25}{10} = \dfrac{5}{2}$$
4) Half the chord length: $\sqrt{r_1^2 - d^2} = \sqrt{9 - \dfrac{25}{4}} = \sqrt{\dfrac{11}{4}} = \dfrac{\sqrt{11}}{2}$.
The diameter of the minimal circle is $\sqrt{11}$, so its circumference is $\pi D = \mathbf{\sqrt{11}\pi}$.
The phrase "minimal area circle passing through two intersections" translates geometrically to: the common chord serves as the diameter. Solve for the chord length to get the diameter immediately.
π― [Self-Diagnosis Variant 02] (Center Coordinates of Minimal Area Circle)
Find the center of the minimal area circle passing through the intersections of $x^2 + y^2 = 8$ and $(x - 2)^2 + (y - 2)^2 = 4$.
View Variant Solution
Breakdown: The center is the midpoint of the common chord, which is the intersection of the line of centers $y = x$ and the Radical Axis $x + y = 3$. Solving yields $\mathbf{(3/2, 3/2)}$.
[Problem 03] Maximum Distance from the Origin to a Parameter-Rotated Radical Axis
Circle $C_1: x^2 + y^2 - 4 = 0$ intersects $C_2: (x - 3\cos\theta)^2 + (y - 3\sin\theta)^2 = 9$. Find the maximum distance from the origin to their common chord as $\theta$ varies.
π Solution & Step-by-Step Breakdown
Answer: $\dfrac{2}{3}$
[Solution]
1) Expand $C_2$: $x^2 + y^2 - 6(\cos\theta)x - 6(\sin\theta)y = 0$.
2) Radical Axis ($C_1 - C_2 = 0$):
$$6(\cos\theta)x + 6(\sin\theta)y - 4 = 0 \implies 3(\cos\theta)x + 3(\sin\theta)y - 2 = 0$$
3) Distance from $(0, 0)$ to the Radical Axis:
$$d = \dfrac{|-2|}{\sqrt{9\cos^2\theta + 9\sin^2\theta}} = \dfrac{2}{\sqrt{9}} = \mathbf{\dfrac{2}{3}}$$
The distance is constant and independent of $\theta$.
When trigonometric coefficients $\cos\theta, \sin\theta$ define center coordinates, the denominator of the distance formula evaluates to a constant via the identity $\cos^2\theta + \sin^2\theta = 1$.
π― [Self-Diagnosis Variant 03] (Radical Axis Invariant Fixed Point)
Find the fixed point through which the common chord of $x^2 + y^2 = 1$ and $x^2 + y^2 - 2kx - 4ky + 2k - 1 = 0$ always passes for any $k \ne 0$.
View Variant Solution
Breakdown: Subtracting gives $2k(x + 2y - 1) = 0 \implies x + 2y = 1$. The chord itself remains the invariant line $\mathbf{x + 2y = 1}$.
[Problem 04] Total Count of Integer Common Chord Lengths
Two circles $C_1: x^2 + y^2 = 25$ and $C_2: (x - 3)^2 + (y - 4)^2 = r^2$ intersect at two distinct points. Find the number of possible integer values for the length of their common chord.
π Solution & Step-by-Step Breakdown
Answer: $9$ values
[Solution]
1) Center distance $d = \sqrt{3^2 + 4^2} = 5$. Radius of $C_1$ is $r_1 = 5$.
2) If $x$ is the perpendicular distance from the center of $C_1$ to the common chord, the chord length is $L = 2\sqrt{25 - x^2}$.
3) For two distinct intersections to exist, $0 < x < 5$.
Thus, $0 < L < 2\sqrt{25} = 10$.
4) Integer values: $L \in \{1, 2, 3, 4, 5, 6, 7, 8, 9\}$, giving exactly $\mathbf{9}$ possible integer lengths.
The length of a common chord can never exceed the diameter of the smaller circle: $0 < L < 2\min(r_1, r_2)$. Establishing this open bound determines the integer count immediately.
π― [Self-Diagnosis Variant 04] (Isosceles Triangle on Common Chord)
Find the area of the triangle formed by the origin and the intersections of $x^2 + y^2 = 16$ and $(x - 4)^2 + y^2 = 16$.
View Variant Solution
Breakdown: Common chord is $x = 2$, so height $d = 2$. Half chord $= \sqrt{16 - 4} = 2\sqrt{3}$, total chord $= 4\sqrt{3}$. Area $= \dfrac{1}{2}(4\sqrt{3})(2) = \mathbf{4\sqrt{3}}$.
Theme 2. Circumference Bisection & Diameter Passages (05 ~ 08)
[Problem 05] Circumference Bisection with Center Constrained to a Line
Circle $C_1: x^2 + y^2 - 2ax - 2ay + 2a^2 - 9 = 0$ has its center on $y = 2x - 3$. If $C_1$ bisects the circumference of $C_2: x^2 + y^2 - 4x + 2y + 1 = 0$, find the constant $a$.
π Solution & Step-by-Step Breakdown
Answer: $a = 3$
[Solution]
1) The circle whose circumference is bisected is $C_2$. Thus, the common chord must pass through the center of $C_2$, which is $(2, -1)$.
2) Subtract to find the Radical Axis ($C_1 - C_2 = 0$):
$$(4 - 2a)x - (2a + 2)y + 2a^2 - 10 = 0$$
3) Substitute $(2, -1)$ into the Radical Axis:
$$(4 - 2a)(2) - (2a + 2)(-1) + 2a^2 - 10 = 0 \implies 2a^2 - 2a = 0 \implies a = 0 \text{ or } a = 1$$
4) Enforcing that the center of $C_1$, which is $(a, a)$, lies on $y = 2x - 3$:
$$a = 2a - 3 \implies a = \mathbf{3}$$
Distinguish between the subject and the object: "Circle $A$ bisects circle $B$" means the center of circle $B$ must satisfy $A - B = 0$. Plugging the coordinates of $B$'s center into the Radical Axis resolves the unknown parameter in one step.
π― [Self-Diagnosis Variant 05] (Circumference Bisection Parameter Relation)
Find the relation between $a$ and $b$ if $x^2 + y^2 + 2ax - 4y + 1 = 0$ bisects the circumference of $x^2 + y^2 - 2x + 2by - 3 = 0$.
View Variant Solution
Breakdown: Substitute the center of the bisected circle $(1, -b)$ into the Radical Axis $(2a + 2)x - (4 + 2b)y + 4 = 0$ to obtain the relation.
[Problem 06] Geometric Impossibility of Mutual Circumference Bisection
Prove whether it is geometrically possible for two distinct circles $C_1$ and $C_2$ to mutually bisect each other's circumferences simultaneously.
π Solution & Step-by-Step Breakdown
Answer: Impossible for distinct circles
[Proof]
1) If $C_1$ bisects $C_2$, the chord is a diameter of $C_2$, so chord length $= 2r_2$.
2) If $C_2$ simultaneously bisects $C_1$, the chord is a diameter of $C_1$, so chord length $= 2r_1$.
3) Equating both yields $r_1 = r_2$.
4) For both bisections to hold, the common chord must pass through both centers $O_1$ and $O_2$. This requires the line of centers $\overline{O_1 O_2}$ to lie directly on the common chord.
However, the line of centers is strictly perpendicular ($90^\circ$) to the common chord. The only way a line can be perpendicular to itself is if $O_1$ and $O_2$ coincide ($d = 0$).
With identical centers and equal radii, the two circles become identical. Therefore, mutual bisection is impossible for distinct circles.
This is a recurring true/false proposition on advanced exams. The perpendicularity condition $\overline{O_1 O_2} \perp \overline{\text{Chord}}$ fundamentally prevents the chord from containing both centers unless the circles are identical.
π― [Self-Diagnosis Variant 06] (Radius Inequality in Circumference Bisection)
If circle $C_1$ bisects the circumference of $C_2$, determine the inequality relating radii $r_1$ and $r_2$.
View Variant Solution
Breakdown: The common chord is a diameter of $C_2$ (length $2r_2$) but merely a secant chord of $C_1$, which cannot exceed $C_1$'s diameter: $2r_2 < 2r_1 \implies \mathbf{r_1 > r_2}$.
[Problem 07] Minimum Radius of a Circle Bisecting Another Circle and Passing Through the Origin
Circle $C_2$ passes through the origin $(0, 0)$ and bisects the circumference of $C_1: (x - 3)^2 + (y - 4)^2 = 4$. Find the minimum radius of $C_2$.
π Solution & Step-by-Step Breakdown
Answer: $\dfrac{\sqrt{29}}{2}$
[Solution]
1) Bisection implies the common chord passes through the center $A(3, 4)$ of $C_1$. The chord endpoints $P, Q$ are diameter endpoints of $C_1$, so $A$ is the midpoint of $PQ$ and $PQ = 2r_1 = 4$.
2) Circle $C_2$ circumscribes $\triangle OPQ$.
3) By the median theorem on right triangles, the circumradius reaches its minimum when segment $OA$ is perpendicular to diameter $PQ$.
With $OA = \sqrt{3^2 + 4^2} = 5$ and $r_1 = 2$, computing the circumradius gives $R_{\min} = \mathbf{\dfrac{\sqrt{29}}{2}}$.
Bisection fixes the midpoint of chord $PQ$ at the center of the smaller circle. Treating the configuration as a circumcircle minimization over three fixed geometric constraints yields the solution via perpendicularity.
π― [Self-Diagnosis Variant 07] (Locus of Centers of Bisecting Circles)
Circle $C$ bisects $x^2 + y^2 = 9$ and passes through $(4, 0)$. Find the equation of the line on which the center of $C$ must lie.
View Variant Solution
Breakdown: Enforcing the power of the point and perpendicular bisector conditions produces the linear locus equation $\mathbf{8x - 25 = 0}$.
[Problem 08] Angle Subtended by Tangents Under Circumference Bisection
Circle $C_1: x^2 + y^2 = 25$ bisects the circumference of $C_2: (x - 3)^2 + (y - 4)^2 = r^2$. Find the angle $\theta$ between the two tangents drawn from the origin $O$ to circle $C_2$.
π Solution & Step-by-Step Breakdown
Answer: $90^\circ$
[Solution]
1) Bisection requires the common chord to pass through center $O_2(3, 4)$ of circle $C_2$.
2) The distance from origin $O_1(0, 0)$ to $O_2(3, 4)$ is $d = \sqrt{3^2 + 4^2} = 5$.
3) Because the radius of $C_1$ is $r_1 = 5$, center $O_2$ lies directly on the perimeter of $C_1$.
4) By the right-triangle tangency relation from the origin to $C_2$, $\sin(\theta/2) = \dfrac{r_2}{d}$. Geometric compatibility locks $\sin(\theta/2) = \dfrac{\sqrt{2}}{2} \implies \theta/2 = 45^\circ \implies \theta = \mathbf{90^\circ}$.
Recognizing that the center $O_2(3, 4)$ lies on the boundary of $C_1$ immediately establishes an isosceles right-triangle configuration, yielding an angle of $90^\circ$ without lengthy algebra.
π― [Self-Diagnosis Variant 08] (Circumference Bisection Center Separation)
Circle $x^2 + y^2 = 16$ bisects the circumference of $(x - a)^2 + y^2 = 4$. Find the positive constant $a$.
View Variant Solution
Breakdown: By Pythagoras on the bisected circle's center: $a = \sqrt{r_1^2 - r_2^2} = \sqrt{16 - 4} = \mathbf{2\sqrt{3}}$.
Theme 3. Circle Pencils & The Radical Center (09 ~ 11)
[Problem 09] Pencil Circle Passing Through Intersections and Tangent to an Axis
Find the radius of the circle that passes through the intersections of $C_1: x^2 + y^2 = 4$ and $C_2: x^2 + y^2 - 4x - 4y + 4 = 0$ and is tangent to the $x$-axis.
π Solution & Step-by-Step Breakdown
Answer: $R = 2$
[Solution]
1) Set up the linear combination (pencil):
$$(x^2 + y^2 - 4) + k(x^2 + y^2 - 4x - 4y + 4) = 0 \quad (k \ne -1)$$
$$(1 + k)x^2 + (1 + k)y^2 - 4kx - 4ky + (4k - 4) = 0$$
2) Normalize coefficients: Center is $(a, b) = \left(\dfrac{2k}{1+k}, \dfrac{2k}{1+k}\right)$.
3) Tangency to the $x$-axis requires radius $R = |b| = \left|\dfrac{2k}{1+k}\right|$.
4) Evaluating $R^2 = a^2 + b^2 - C$ gives $k = 1$, yielding radius $R = \mathbf{2}$.
Use the pencil formulation $C_1 + kC_2 = 0$ to express the center coordinates in terms of $k$, then enforce the axis tangency constraint $|b| = R$ to solve for $k$ algebraically.
π― [Self-Diagnosis Variant 09] (Pencil Circle Tangent to $y$-Axis)
Find the $x$-coordinate of the center of a circle passing through the intersections of $x^2 + y^2 - 2x = 0$ and $x^2 + y^2 - 4y = 0$ that is tangent to the $y$-axis.
View Variant Solution
Breakdown: Apply the $y$-axis tangency condition $|a| = R$ to the normalized pencil coordinates to find $x = \mathbf{1}$.
[Problem 10] Locus of Centers of Coaxial Circles
Find the equation of the locus of centers $P(x, y)$ of all circles passing through the intersections of $x^2 + y^2 = 9$ and $x^2 + y^2 - 6x - 8y + 21 = 0$.
π Solution & Step-by-Step Breakdown
Answer: $4x - 3y = 0$
[Solution]
1) Every circle in a coaxial pencil shares the same Radical Axis.
2) The perpendicular bisector of this common chord is precisely the line of centers connecting $O_1(0, 0)$ and $O_2(3, 4)$.
3) The equation of the line passing through $(0, 0)$ and $(3, 4)$ is:
$$y = \dfrac{4}{3}x \implies \mathbf{4x - 3y = 0}$$
The locus of centers of all circles passing through two common intersection points is always the line of centers of the two generating circles.
π― [Self-Diagnosis Variant 10] (Slope of the Line of Centers)
Find the slope of the line of centers for circles passing through the intersections of $(x - 1)^2 + (y - 2)^2 = 4$ and $(x - 5)^2 + (y - 4)^2 = 9$.
View Variant Solution
Breakdown: Slope between centers $(1, 2)$ and $(5, 4)$ is $\dfrac{4 - 2}{5 - 1} = \mathbf{\dfrac{1}{2}}$.
[Problem 11] Concurrence of Three Radical Axes (The Radical Center)
For circles $C_1: x^2 + y^2 = 4$, $C_2: x^2 + y^2 - 4x = 0$, and $C_3: x^2 + y^2 - 2y - k = 0$, prove that their three pairwise Radical Axes concur at a single point for any real constant $k$.
π Solution & Step-by-Step Breakdown
Answer: Concurs for all real $k$ (Identity holds)
[Proof]
1) Radical Axis of $C_1$ and $C_2$: $C_1 - C_2 = 0 \implies 4x - 4 = 0 \implies x = 1$.
2) Radical Axis of $C_1$ and $C_3$: $C_1 - C_3 = 0 \implies 2y + k - 4 = 0 \implies y = \dfrac{4 - k}{2}$.
3) Intersection of the first two axes: $\left(1, \dfrac{4-k}{2}\right)$.
4) Radical Axis of $C_2$ and $C_3$: $C_2 - C_3 = 0 \implies -4x + 2y + k = 0$.
Substitute the intersection coordinates:
$$-4(1) + 2\left(\dfrac{4 - k}{2}\right) + k = -4 + 4 - k + k = 0$$
This identity holds for all $k \in \mathbb{R}$. The concurrent point is the Radical Center.
The linear dependency $(C_1 - C_2) + (C_2 - C_3) = C_1 - C_3$ guarantees that whenever three circle centers are non-collinear, their three Radical Axes concur at a unique point called the Radical Center.
π― [Self-Diagnosis Variant 11] (Locating the Radical Center)
Find the coordinates of the Radical Center of $x^2 + y^2 = 1$, $x^2 + y^2 - 2x = 0$, and $x^2 + y^2 - 4y = 0$.
View Variant Solution
Breakdown: $C_1 - C_2 = 0 \implies x = 1/2$. $C_1 - C_3 = 0 \implies y = 1/4$. Radical Center is $\mathbf{(1/2, 1/4)}$.
Theme 4. Power of a Point & Orthogonal Intersections (12 ~ 13)
[Problem 12] Invariance of Tangent Segment Lengths from the Radical Axis
From a point $P(x_0, y_0)$ on the extension of the common chord of $C_1: x^2 + y^2 = 9$ and $C_2: (x - 4)^2 + y^2 = 1$, tangents of lengths $L_1$ and $L_2$ are drawn to $C_1$ and $C_2$. Find the ratio $L_1 : L_2$.
π Solution & Step-by-Step Breakdown
Answer: $1 : 1$
[Solution]
1) By the Power of a Point Theorem, tangent segment lengths squared are:
$$L_1^2 = PO_1^2 - r_1^2 = C_1(x_0, y_0)$$
$$L_2^2 = PO_2^2 - r_2^2 = C_2(x_0, y_0)$$
2) Point $P$ lies on the Radical Axis, which is defined by $C_1(x, y) - C_2(x, y) = 0$.
Therefore, $C_1(x_0, y_0) = C_2(x_0, y_0) \implies L_1^2 = L_2^2 \implies L_1 = L_2$.
The ratio is always $\mathbf{1 : 1}$.
The geometric definition of the Radical Axis is the locus of all points having equal power with respect to both circles. Thus, tangent segment lengths drawn from anywhere along the Radical Axis are always equal.
π― [Self-Diagnosis Variant 12] (Calculating Tangent Length from Axis Intercept)
Find the length of the tangents to $x^2 + y^2 = 16$ and $(x - 6)^2 + y^2 = 4$ drawn from the $y$-intercept of their common chord.
View Variant Solution
Breakdown: Compute the $y$-intercept of $C_1 - C_2 = 0$, then evaluate $L = \sqrt{d^2 - 16}$.
[Problem 13] Common Chord Length of Orthogonal Circles
Two circles $C_1: x^2 + y^2 = 16$ and $C_2: (x - 5)^2 + y^2 = 9$ intersect orthogonally (their tangents at each intersection point are perpendicular). Find the length of their common chord.
π Solution & Step-by-Step Breakdown
Answer: $\dfrac{24}{5}$
[Solution]
1) Orthogonality means the radii to intersection $A$ meet at $90^\circ$: $\triangle O_1 A O_2$ is a right triangle with $\angle O_1 A O_2 = 90^\circ$.
2) Legs: $r_1 = 4, r_2 = 3$. Hypotenuse: $d = O_1 O_2 = 5$. ($3-4-5$ right triangle).
3) The common chord is twice the altitude $AH$ to hypotenuse $O_1 O_2$.
By equating area products:
$$AH = \dfrac{r_1 \cdot r_2}{d} = \dfrac{4 \cdot 3}{5} = \dfrac{12}{5}$$
Total chord length $= 2 \times AH = \mathbf{\dfrac{24}{5}}$.
Two circles are orthogonal if and only if $d^2 = r_1^2 + r_2^2$. In this setting, the chord length is given directly by the altitude formula: $\text{Chord} = \dfrac{2r_1 r_2}{d}$.
π― [Self-Diagnosis Variant 13] (Orthogonal Chord Length with 5-12-13 Triangle)
Find the common chord length of two orthogonal circles with radii $5$ and $12$ and center distance $13$.
View Variant Solution
Breakdown: Since $5^2 + 12^2 = 13^2$, the circles are orthogonal: $\text{Chord} = \dfrac{2(5)(12)}{13} = \mathbf{\dfrac{120}{13}}$.
Theme 5. Common Tangents & Radical Axis Enclosures (14 ~ 15)
[Problem 14] Distance Between Intersections of Common Tangents and the Radical Axis
Find the distance between the two points where the two direct common tangents intersect the extension of the common chord of $C_1: x^2 + y^2 = 4$ and $C_2: (x - 6)^2 + y^2 = 16$.
π Solution & Step-by-Step Breakdown
Answer: $4\sqrt{2}$
[Solution]
1) Radical Axis ($C_1 - C_2 = 0$):
$$(x^2 + y^2 - 4) - (x^2 + y^2 - 12x + 20) = 0 \implies 12x - 24 = 0 \implies x = 2$$
2) The direct common tangents meet at the external homothetic center $P(-6, 0)$, dividing the centers $O_1(0, 0)$ and $O_2(6, 0)$ in the ratio $r_1 : r_2 = 1 : 2$.
3) Slopes of direct common tangents from $P(-6, 0)$:
$$\dfrac{|6m|}{\sqrt{m^2 + 1}} = 2 \implies 32m^2 = 4 \implies m = \pm \dfrac{\sqrt{2}}{4}$$
4) Intersections with $x = 2$: $y = \pm \dfrac{\sqrt{2}}{4}(2 + 6) = \pm 2\sqrt{2}$.
Distance between points: $2\sqrt{2} - (-2\sqrt{2}) = \mathbf{4\sqrt{2}}$.
The Radical Axis is perpendicular to the line of centers. Consequently, its intersection points with symmetric direct common tangents are reflection-symmetric across the line of centers.
π― [Self-Diagnosis Variant 14] (Transverse Tangents and Radical Axis Concurrence)
Find the distance between the intersection points of the two transverse common tangents and the common chord of $x^2 + y^2 = 1$ and $(x - 4)^2 + y^2 = 1$.
View Variant Solution
Breakdown: With equal radii, the internal homothetic center $(2, 0)$ lies exactly on the Radical Axis $x = 2$, meaning all three lines concur at a single point (distance $= \mathbf{0}$).
[Problem 15] Area of Rhombus Formed by Common Chord and Centers
Find the area of the rhombus formed by the common chord and the two centers of circles $C_1: x^2 + y^2 = 25$ and $C_2: (x - 8)^2 + y^2 = 25$.
π Solution & Step-by-Step Breakdown
Answer: $24$
[Solution]
1) The horizontal diagonal connects centers $O_1(0, 0)$ and $O_2(8, 0)$, so $d_1 = 8$.
2) The vertical diagonal is the common chord along the line of symmetry $x = 4$.
3) In $C_1: x^2 + y^2 = 25$, substitute $x = 4$:
$$4^2 + y^2 = 25 \implies y^2 = 9 \implies y = \pm 3$$
The vertical diagonal length is $d_2 = 3 - (-3) = 6$.
4) Because the line of centers and the common chord are perpendicular, the rhombus area is:
$$\text{Area} = \dfrac{1}{2} \cdot d_1 \cdot d_2 = \dfrac{1}{2}(8)(6) = \mathbf{24}$$
For two circles with equal radii, the four points consisting of the two centers and the two intersection points form a rhombus. Since diagonals are perpendicular, calculate $\dfrac{1}{2} d_1 d_2$ directly.
π― [Self-Diagnosis Variant 15] (Rhombus Area with Radius $16$)
Find the area of the quadrilateral formed by the common chord and direct common tangent contact points of $x^2 + y^2 = 16$ and $(x - 6)^2 + y^2 = 16$.
View Variant Solution
Breakdown: Common chord length is $2\sqrt{16 - 3^2} = 2\sqrt{7}$, horizontal distance is $6$. Area $= \dfrac{1}{2}(6)(2\sqrt{7}) = \mathbf{6\sqrt{7}}$.
Across our exploration of circle equations, one fundamental geometric truth stands out:
"No matter how complex a problem appears, dropping perpendicular auxiliary radii from the centers reduces everything to pure Pythagorean geometry."
Keep the linear Radical Axis ($C_1 - C_2 = 0$) and the center-passing condition for circumference bisection at the core of your problem-solving toolkit.
With these principles mastered, you are fully prepared to approach the next chapter—[Geometric Transformations: Translations & Reflections]—with clarity and confidence.
Comments
Post a Comment