JEE Main & Advanced

Circles

Coordinate Geometry: Circles

1
Module 1

Center-Radius Forms & Core Configurations

Standard, General, Diameter, and Parametric FormsTopic 1

A circle is geometrically defined as the locus of a variable point that moves in a two-dimensional plane such that its distance from a fixed point (the center) remains constantly fixed (the radius).
  • Standard Form: A circle centered at $H(h, k)$ with radius $r$ satisfies: \[ (x-h)^2 + (y-k)^2 = r^2 \]
  • General Form: Written as $x^2 + y^2 + 2gx + 2fy + c = 0$. By completing the square, the central characteristics are extracted directly from the coefficients: \[ \text{Center: } (-g, -f), \quad \text{Radius: } \sqrt{g^2 + f^2 - c} \]
    1. If $g^2 + f^2 - c > 0$, the circle is a real circle.
    2. If $g^2 + f^2 - c = 0$, it collapses into a point circle at $(-g, -f)$.
    3. If $g^2 + f^2 - c < 0$, the radius is imaginary, defining a virtual curve.
  • Diameter Form: If $A(x_1, y_1)$ and $B(x_2, y_2)$ form the endpoints of a diameter segment, any point $P(x,y)$ on the boundary forms a right angle $\angle APB = 90^\circ$. Equating the orthogonal slope product ($m_{AP} \cdot m_{BP} = -1$) yields the joint equation: \[ (x - x_1)(x - x_2) + (y - y_1)(y - y_2) = 0 \]
  • Parametric Form: Expresses coordinates dynamically using an angular parameter $\theta$: \[ x = h + r\cos\theta, \quad y = k + r\sin\theta \quad (0 \le \theta < 2\pi) \]
Worked Examples
1

Find the center coordinates and radius of the circle represented by the general second-degree equation $2x^2 + 2y^2 - 8x + 12y - 10 = 0$.

Show solution

First, normalize the equation by dividing all terms by $2$ so that the leading quadratic coefficients match unity exactly: \[ x^2 + y^2 - 4x + 6y - 5 = 0 \] Extract the general coefficients by comparing with the standard form: \[ 2g = -4 \implies g = -2, \quad 2f = 6 \implies f = 3, \quad c = -5 \] Compute the center parameters: \[ \text{Center: } (-g, -f) = (-(-2), -3) = (2, -3) \] Compute the metric radius: \[ r = \sqrt{g^2 + f^2 - c} = \sqrt{(-2)^2 + (3)^2 - (-5)} = \sqrt{4 + 9 + 5} = \sqrt{18} = 3\sqrt{2} \] Final Answer: Center = $(2, -3)$, Radius = $3\sqrt{2}$ units.

✎ Self-Check — 5 questions0 / 5
Q1.Find the center and radius of the circle equation $x^2 + y^2 + 4x - 8y - 5 = 0$:
Q2.Find the equation of the circle having diameter endpoints at $A(1, 2)$ and $B(3, 4)$:
Q3.The parametric tracking trajectory $x = -1 + 3\cos\theta$, $y = 2 + 3\sin\theta$ maps to which general Cartesian circle form?
Q4.For what value of the constant parameter $c$ does the general expression $x^2 + y^2 - 6x + 8y + c = 0$ represent a point circle?
Q5.If a straight line intercept circle cuts the axes at $(a, 0)$ and $(0, b)$ and passes through the origin $(0,0)$, its diameter form equation is:
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Module 2

Tangency Metrics & Point Transformations

Point Power and Tangent/Normal Line EquationsTopic 1

Let $S \equiv x^2 + y^2 + 2g|x + 2fy + c = 0$ represent our baseline circle. Evaluating a point $P(x_1, y_1)$ inside the expression defines the Power of a Point, denoted as $S_1 = x_1^2 + y_1^2 + 2gx_1 + 2fy_1 + c$. The position of $P$ relative to the circle depends on the sign of $S_1$:
  • $S_1 > 0 \implies P$ is located outside the circle boundary.
  • $S_1 = 0 \implies P$ lies exactly on the circle boundary trace.
  • $S_1 < 0 \implies P$ is situated inside the circle boundary.
If $P$ is outside the circle ($S_1 > 0$), the geometric length $L$ of the tangent line segment drawn from $P$ to the circle is: \[ L = \sqrt{S_1} \] To construct linear boundary lines at a point on the circle, apply standard calculus transformations ($x^2 \to x x_1$, $2x \to x + x_1$, etc.) using the $T = 0$ notation:
  • Tangent Line ($T=0$): At a point $P(x_1, y_1)$ on the circle, the tangent equation is: \[ T \equiv x x_1 + y y_1 + g(x + x_1) + f(y + y_1) + c = 0 \]
  • Slope Form Tangent: For a given slope $m$, a line parallel to $y = mx$ is tangent to the circle $(x-h)^2 + (y-k)^2 = r^2$ if its perpendicular distance from the center equals the radius, yielding: \[ y - k = m(x - h) \pm r\sqrt{1 + m^2} \]
  • Normal Line: A normal line stands perpendicular to the tangent at the point of contact. Since a circle is perfectly symmetric, every normal line must pass through the center coordinates $(-g, -f)$. Thus, the normal equation connecting the center and $P(x_1, y_1)$ is simply the line passing through those two points.
Worked Examples
1

Find the equation of the tangent line to the circle $x^2 + y^2 - 4x + 2y - 5 = 0$ at the point $P(5, 0)$ lying on its boundary.

Show solution

First, verify that $P(5, 0)$ lies on the circle boundary by checking that $S_1 = 0$: \[ 5^2 + 0^2 - 4(5) + 2(0) - 5 = 25 + 0 - 20 + 0 - 5 = 0 \quad (\text{Valid Boundary Point}) \] Apply the $T = 0$ template transformation rules directly: \[ x^2 \to 5x, \quad y^2 \to 0y, \quad 2x \to x + 5, \quad 2y \to y + 0 \] Substitute these into the general equation: \[ 5x + 0y - 2(x + 5) + 1(y + 0) - 5 = 0 \] Expand and gather the variable coefficients: \[ 5x - 2x - 10 + y - 5 = 0 \implies 3x + y - 15 = 0 \] Final Answer: The tangent line equation is $3x + y - 15 = 0$.

✎ Self-Check — 5 questions0 / 5
Q1.Calculate the length of the tangent line segment drawn from the external point $P(6, 5)$ to the circle curve $x^2 + y^2 - 2x - 4y - 4 = 0$:
Q2.The line $y = mx + c$ is tangent to the circle $x^2 + y^2 = R^2$ if the constant parameter satisfies which condition?
Q3.Find the equation of the normal line to the circle $x^2 + y^2 - 6x - 4y - 12 = 0$ at the boundary point $P(6, 6)$:
Q4.Determine the structural position of the point $Q(1, 2)$ relative to the circle curve $x^2 + y^2 - 4x - 2y + 2 = 0$:
Q5.The equation of the tangent line to the circle $x^2 + y^2 = 25$ at the point $(3, 4)$ is:

Chords of Contact and Midpoint Structural ParametrizationsTopic 2

This subsection covers chord lines generated by tracking points and lines across the circle field:
  • Chord of Contact: If two tangent lines are drawn from an external point $P(x_1, y_1)$ to a circle, they touch the boundary at two points of tangency, $A$ and $B$. The line segment $AB$ connecting these points is the chord of contact. Its equation matches the $T = 0$ format: \[ x x_1 + y y_1 + g(x + x_1) + f(y + y_1) + c = 0 \] Note that while this equation shares the same format as a tangent line, it represents a chord line because $P$ lies outside the circle boundary.
  • Chord with a Given Midpoint: If a chord line is bisected exactly at an interior point $M(x_1, y_1)$, its equation can be found using the $T = S_1$ formula: \[ x x_1 + y y_1 + g(x + x_1) + f(y + y_1) + c = x_1^2 + y_1^2 + 2gx_1 + 2fy_1 + c \]
Worked Examples
1

Find the equation of the chord of the circle $x^2 + y^2 = 25$ whose midpoint is located at $M(2, -1)$.

Show solution
Apply the $T = S_1$ chord formula directly.
  • Step 1: Formulate the $T$ expression mapping
    \[ T = x(2) + y(-1) = 2x - y \]
  • Step 2: Evaluate the $S_1$ scalar power value
    \[ S_1 = (2)^2 + (-1)^2 = 4 + 1 = 5 \]
Equate the two expressions ($T = S_1$): \[ 2x - y = 5 \implies 2x - y - 5 = 0 \] Final Answer: The chord equation is $2x - y - 5 = 0$.
✎ Self-Check — 5 questions0 / 5
Q1.Find the equation of the chord of contact drawn from the external point $P(4, 3)$ to the circle curve $x^2 + y^2 = 9$:
Q2.Find the equation of the chord of the circle $x^2 + y^2 - 4x - 6y - 3 = 0$ that is bisected exactly at the point $M(1, 1)$:
Q3.If the chord of contact of tangents drawn from a variable point $P$ to the circle $x^2 + y^2 = R^2$ always passes through a fixed point $(a, b)$, then the locus of $P$ forms a:
Q4.The length of a chord of a circle with radius $R$ whose midpoint is located at a distance $d$ from the center point evaluates to:
Q5.Find the equation of the chord of contact from the origin $(0,0)$ to the circle $(x-2)^2 + (y-3)^2 = 4$:
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Module 3

Multi-Circle Operators & Radical Fields

Radical Axis, Centres, and Common ChordsTopic 1

The geometric relationships between multiple circles can be analyzed using linear combinations of their equations:
  • Radical Axis: The radical axis of two circles $S_1 = 0$ and $S_2 = 0$ is the locus of a point that moves such that the lengths of the tangents drawn from it to both circles are equal. Mathematically, it is found by subtracting the two normalized equations: \[ S_1 - S_2 = 0 \implies 2(g_1 - g_2)x + 2(f_1 - f_2)y + (c_1 - c_2) = 0 \] The radical axis is always a straight line perpendicular to the line joining the centers of the two circles. If the circles intersect, the radical axis matches their common chord line.
  • Radical Center: For a system of three circles, the three independent radical axes intersect at a single point called the radical center. The lengths of the tangents drawn from this point to all three circles are equal.
Worked Examples
1

Find the equation of the common chord (radical axis) for the two circles $S_1: x^2 + y^2 - 4x - 6y + 3 = 0$ and $S_2: x^2 + y^2 - 2x - 2y - 1 = 0$.

Show solution

Verify that both circle equations are normalized, then subtract $S_2$ from $S_1$: \[ S_1 - S_2 = 0 \] \[ (x^2 + y^2 - 4x - 6y + 3) - (x^2 + y^2 - 2x - 2y - 1) = 0 \] Combine like terms to find the linear equation: \[ -4x + 2x - 6y + 2y + 3 + 1 = 0 \implies -2x - 4y + 4 = 0 \] Divide by $-2$ to simplify the coefficients: \[ x + 2y - 2 = 0 \] Final Answer: The common chord equation is $x + 2y - 2 = 0$.

✎ Self-Check — 5 questions0 / 5
Q1.Find the equation of the radical axis for the two circles $x^2 + y^2 + 2x + 4y + 1 = 0$ and $x^2 + y^2 + 4x + 2y + 1 = 0$:
Q2.The radical axis of two non-intersecting circles stands perpendicular to:
Q3.Find the length of the common chord of the intersecting circles $x^2 + y^2 = 4$ and $x^2 + y^2 - 4x = 0$:
Q4.The radical center of three mutually intersecting circles where each pair has a common chord line is the point of intersection of their:
Q5.If two circles touch each other externally, their radical axis matches their:

Intersection Angles, Orthogonality, and Coaxial SystemsTopic 2

Advanced multi-circle architectures track intersection profiles and system configurations:
  • Angle of Intersection: The angle $\phi$ between two intersecting circles is defined as the angle between their tangent lines at a point of intersection. It can be found from the radii and center distance $d$ using the Law of Cosines: \[ \cos\phi = \frac{r_1^2 + r_2^2 - d^2}{2r_1 r_2} \]
  • Orthogonal Circles ($\phi = 90^\circ$): Two circles are orthogonal if they intersect at a right angle. Setting $\cos(90^\circ) = 0$ simplifies the relationship to $d^2 = r_1^2 + r_2^2$. Written in terms of the general coefficients, this yields the standard orthogonality condition: \[ 2g_1 g_2 + 2f_1 f_2 = c_1 + c_2 \]
  • Coaxial System of Circles: A system of circles is coaxial if every pair of circles in the system shares the exact same radical axis. The family can be generated using a base circle $S$ and the common radical axis line $L$: \[ S + \lambda L = 0 \]
Worked Examples
1

Find the value of the constant parameter $k$ for which the two circles $x^2 + y^2 - 4x + 6y + 5 = 0$ and $x^2 + y^2 + kx - 2y + 9 = 0$ intersect orthogonally.

Show solution
Extract the general coefficients from both circle equations:
  • Circle 1: $2g_1 = -4 \implies g_1 = -2, \quad 2f_1 = 6 \implies f_1 = 3, \quad c_1 = 5$
  • Circle 2: $2g_2 = k \implies g_2 = \frac{k}{2}, \quad 2f_2 = -2 \implies f_2 = -1, \quad c_2 = 9$
Apply the standard orthogonality condition equation: \[ 2g_1 g_2 + 2f_1 f_2 = c_1 + c_2 \] Substitute the extracted coefficients into the formula: \[ 2(-2)\left(\frac{k}{2}\right) + 2(3)(-1) = 5 + 9 \] Simplify the terms and solve for $k$: \[ -2k - 6 = 14 \implies -2k = 20 \implies k = -10 \] Final Answer: The circles are orthogonal when $k = -10$.
✎ Self-Check — 5 questions0 / 5
Q1.Find the value of $c$ if the circles $x^2 + y^2 - 2x - 4y + c = 0$ and $x^2 + y^2 - 4x - 2y + 4 = 0$ cut orthogonally:
Q2.If two circles are orthogonal, the angle of intersection between their boundaries is exactly:
Q3.Any line belonging to the family of radical axes inside a coaxial system of circles is:
Q4.Find the angle of intersection between the two circles $x^2 + y^2 = 4$ and $x^2 + y^2 - 4x = 0$ if their center distance matches $d = 2$:
Q5.The condition for the two general circles to be orthogonal depends strictly on which coefficients?
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Module 4

Tangent Topology & Relative Configurations

Relative Positions and Common Tangent ProfilesTopic 1

The relative position of two circles in a plane depends on the relationship between their center distance $d$ and the sum or difference of their radii, $r_1$ and $r_2$. This relationship also determines the number of common tangent lines that can be drawn to the circles:
  • Case 1: Circles are completely separate ($d > r_1 + r_2$)
    The circles do not intersect or touch. There are 4 common tangents: 2 direct (external) tangents and 2 transverse (internal) tangents.
  • Case 2: Circles touch externally ($d = r_1 + r_2$)
    There are 3 common tangents: 2 direct tangents and 1 common internal tangent at the point of contact.
  • Case 3: Circles intersect at two points ($|r_1 - r_2| < d < r_1 + r_2$)
    There are 2 common tangents: both are direct external tangents.
  • Case 4: Circles touch internally ($d = |r_1 - r_2|$)
    There is 1 common tangent: a single common external tangent passing through the point of contact.
  • Case 5: One circle lies entirely inside the other ($d < |r_1 - r_2|$)
    The circles do not touch or intersect. There are 0 common tangents.
Direct tangents intersect on the line of centers at an external point $O_1$ that divides the center segment externally in the ratio of the radii ($r_1 : r_2$). Transverse tangents intersect at an internal point $O_2$ that divides the center segment internally in the same ratio ($r_1 : r_2$).
Worked Examples
1

Determine the relative position and the total number of common tangents for the two circles $S_1: x^2 + y^2 = 4$ and $S_2: x^2 + y^2 - 6x - 8y + 21 = 0$.

Show solution
Extract the center coordinates and radii for both circles:
  • Circle 1: Center $C_1(0, 0)$, Radius $r_1 = 2$
  • Circle 2: Center $C_2(3, 4)$, Radius $r_2 = \sqrt{(-3)^2 + (-4)^2 - 21} = \sqrt{9 + 16 - 21} = \sqrt{4} = 2$
Compute the distance $d$ between the centers $C_1$ and $C_2$: \[ d = \sqrt{(3 - 0)^2 + (4 - 0)^2} = \sqrt{9 + 16} = \sqrt{25} = 5 \] Evaluate the sum of the radii: \[ r_1 + r_2 = 2 + 2 = 4 \] Compare the center distance with the sum of the radii: \[ d = 5 > 4 = r_1 + r_2 \] Since the center distance is strictly greater than the sum of the radii ($d > r_1 + r_2$), the circles are completely separate and do not touch or intersect. This configuration allows for the maximum number of common tangents. Final Answer: The circles are separate; there are exactly 4 common tangents.
✎ Self-Check — 5 questions0 / 5
Q1.If the center distance between two circles satisfies $d = r_1 + r_2$, the total number of common tangents that can be drawn is:
Q2.Find the total number of common tangents for the two circles $x^2 + y^2 = 9$ and $x^2 + y^2 - 2x = 0$ if they intersect at two points:
Q3.The internal intersection point $O_2$ of the transverse common tangents divides the line segment joining the centers of the circles in what ratio?
Q4.If the center distance satisfies $d < |r_1 - r_2|$, the total number of common tangents matches:
Q5.Find the relative position condition if two circles satisfy the center-radius equation $d = |r_1 - r_2|$:

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