The Hook: The Circular World Around Us
Look around you right now:
- The clock on your wall showing time in a circular path
- Your coffee mug with a circular rim
- Wheels of vehicles, steering wheels, satellite dishes, Ferris wheels
From “Pathaan’s” Car Chase: When Shah Rukh Khan drifts a car around a circular curve, the physics involves circular motion. But how do we describe that circular path mathematically?
The Question: How do we write the equation of a circle? How do engineers calculate where a road should be tangent to a circular roundabout?
JEE Impact: Circles appear in 4-5 questions every year in JEE Main and 2-3 in Advanced. This is one of the highest-weightage topics in coordinate geometry!
The Core Concept
A circle is the set of all points at a fixed distance (radius) from a fixed point (center).
The Big Idea
Think of drawing a circle with a compass:
- You fix the pointed end at one location (center)
- The pencil traces all points at the same distance (radius)
In coordinates: If center is $(h, k)$ and radius is $r$, then for any point $(x, y)$ on the circle:
$$\text{Distance from }(x,y)\text{ to }(h,k) = r$$This simple idea gives us the circle equation!
Interactive Demo: Explore Circle Equations
Plot different circle equations and see how changing the center $(h, k)$ and radius $r$ affects the graph! Try circles like $(x-2)^2 + (y-3)^2 = 25$ to visualize.
Standard Form of Circle
$$\boxed{(x - h)^2 + (y - k)^2 = r^2}$$Where:
- $(h, k)$ = center of the circle
- $r$ = radius of the circle
Special Case - Circle at Origin:
$$\boxed{x^2 + y^2 = r^2}$$Center: $(0, 0)$, Radius: $r$
A Ferris wheel has its center at point $(3, 4)$ and radius $5$ meters. What’s its equation?
Solution:
$$(x - 3)^2 + (y - 4)^2 = 25$$To check if a cabin at $(6, 8)$ is on the wheel:
$$(6-3)^2 + (8-4)^2 = 9 + 16 = 25$$✓ Yes, it’s on the circle!
General Form of Circle
$$\boxed{x^2 + y^2 + 2gx + 2fy + c = 0}$$Where:
- Center: $(-g, -f)$
- Radius: $r = \sqrt{g^2 + f^2 - c}$
For a real circle to exist:
$$\boxed{g^2 + f^2 - c > 0}$$- If $g^2 + f^2 - c = 0$ → Point circle (radius = 0)
- If $g^2 + f^2 - c < 0$ → Imaginary circle (doesn’t exist in real plane)
Converting Between Forms
Standard to General:
$(x-h)^2 + (y-k)^2 = r^2$
Expand: $x^2 - 2hx + h^2 + y^2 - 2ky + k^2 = r^2$
Rearrange: $x^2 + y^2 - 2hx - 2ky + (h^2 + k^2 - r^2) = 0$
Compare: $g = -h$, $f = -k$, $c = h^2 + k^2 - r^2$
General to Standard:
From $x^2 + y^2 + 2gx + 2fy + c = 0$:
Complete the square:
$$(x+g)^2 - g^2 + (y+f)^2 - f^2 + c = 0$$ $$(x+g)^2 + (y+f)^2 = g^2 + f^2 - c$$Center: $(-g, -f)$, Radius: $\sqrt{g^2 + f^2 - c}$
“Go (g) and Find (f) the Center with MINUS”
In $x^2 + y^2 + 2gx + 2fy + c = 0$:
Center = $(-g, -f)$ ← Notice the minus signs!
Why “2g” and “2f”? So that when we complete the square, we get nice expressions without fractions.
Equation of Circle from Different Conditions
1. Circle with Given Center and Radius
Given: Center $(h, k)$ and radius $r$
Equation: $(x-h)^2 + (y-k)^2 = r^2$
2. Circle with Given Center, Passing Through a Point
Given: Center $(h, k)$ and point $(x_1, y_1)$ on circle
Equation: $(x-h)^2 + (y-k)^2 = (x_1-h)^2 + (y_1-k)^2$
(Right side calculates radius squared)
3. Circle with Diameter Endpoints
Given: Diameter has endpoints $(x_1, y_1)$ and $(x_2, y_2)$
Equation:
$$\boxed{(x - x_1)(x - x_2) + (y - y_1)(y - y_2) = 0}$$Visual Description: If you know the two ends of a diameter, any point on the circle makes a right angle with those endpoints (angle in semicircle = 90°).
Find equation of circle with diameter endpoints $(1, 2)$ and $(3, 4)$.
Solution:
$$(x-1)(x-3) + (y-2)(y-4) = 0$$ $$x^2 - 4x + 3 + y^2 - 6y + 8 = 0$$ $$\boxed{x^2 + y^2 - 4x - 6y + 11 = 0}$$4. Circle Through Three Points
Given: Three non-collinear points $(x_1, y_1)$, $(x_2, y_2)$, $(x_3, y_3)$
Method:
- Assume general form: $x^2 + y^2 + 2gx + 2fy + c = 0$
- Substitute all three points to get three equations
- Solve for $g$, $f$, and $c$
JEE Trick: Use determinant method or family of circles (more efficient!)
Parametric Equations of Circle
For circle $x^2 + y^2 = r^2$:
$$\boxed{x = r\cos\theta, \quad y = r\sin\theta}$$where $\theta$ is the parameter (angle from positive x-axis).
For general circle $(x-h)^2 + (y-k)^2 = r^2$:
$$\boxed{x = h + r\cos\theta, \quad y = k + r\sin\theta}$$Visual Description: As $\theta$ varies from $0$ to $2\pi$, the point $(x, y)$ traces the complete circle once.
Why Parametric? Useful for:
- Finding coordinates of any point on circle
- Solving problems involving motion on circular paths
- Simplifying integration problems
Find a point on circle $x^2 + y^2 = 25$ at angle $60°$ from x-axis.
Solution:
Radius $r = 5$, $\theta = 60°$
$$x = 5\cos 60° = 5 \times \frac{1}{2} = \frac{5}{2}$$ $$y = 5\sin 60° = 5 \times \frac{\sqrt{3}}{2} = \frac{5\sqrt{3}}{2}$$Point: $\left(\frac{5}{2}, \frac{5\sqrt{3}}{2}\right)$
Tangent to a Circle
A tangent is a line that touches the circle at exactly one point.
Visual Description: Think of a wheel touching the road - the road is tangent to the wheel, touching at one point.
Equation of Tangent at Point $(x_1, y_1)$
For circle $x^2 + y^2 = r^2$:
$$\boxed{xx_1 + yy_1 = r^2}$$For circle $(x-h)^2 + (y-k)^2 = r^2$:
$$\boxed{(x-h)(x_1-h) + (y-k)(y_1-k) = r^2}$$For general circle $x^2 + y^2 + 2gx + 2fy + c = 0$:
$$\boxed{xx_1 + yy_1 + g(x+x_1) + f(y+y_1) + c = 0}$$“Square becomes Product, Single stays Single”
To get tangent at $(x_1, y_1)$:
- $x^2$ becomes $xx_1$
- $y^2$ becomes $yy_1$
- $x$ becomes $\frac{x+x_1}{2}$ (or write as $x$ in numerator, $x_1$ in position)
- $y$ becomes $\frac{y+y_1}{2}$
- Constant stays same
Example: For $x^2 + y^2 = 25$, tangent at $(3,4)$:
$x(3) + y(4) = 25$ → $3x + 4y = 25$
Tangent with Given Slope
Equation of tangent to circle $x^2 + y^2 = r^2$ with slope $m$:
$$\boxed{y = mx \pm r\sqrt{1 + m^2}}$$Note: The $\pm$ gives two tangents (one on each side of circle) with the same slope.
For general circle $(x-h)^2 + (y-k)^2 = r^2$:
$$\boxed{y - k = m(x - h) \pm r\sqrt{1 + m^2}}$$Find equations of tangents to $x^2 + y^2 = 25$ parallel to line $3x + 4y = 0$.
Solution:
Slope of given line: $m = -\frac{3}{4}$
Tangents with this slope:
$$y = -\frac{3}{4}x \pm 5\sqrt{1 + \frac{9}{16}}$$ $$y = -\frac{3}{4}x \pm 5\sqrt{\frac{25}{16}}$$ $$y = -\frac{3}{4}x \pm \frac{25}{4}$$Multiply by 4:
$$4y = -3x \pm 25$$ $$\boxed{3x + 4y = 25 \quad \text{and} \quad 3x + 4y = -25}$$Length of Tangent from External Point
Length of tangent from point $(x_1, y_1)$ to circle $x^2 + y^2 + 2gx + 2fy + c = 0$:
$$\boxed{L = \sqrt{x_1^2 + y_1^2 + 2gx_1 + 2fy_1 + c}}$$In simple terms: “Just substitute the point coordinates in the circle equation and take square root”
Visual Description: If you stand outside a circular pond and stretch ropes to touch it tangentially, the length of each rope is given by this formula.
Find length of tangent from $(5, 1)$ to circle $x^2 + y^2 + 4x - 6y - 3 = 0$.
Solution:
$$L = \sqrt{5^2 + 1^2 + 4(5) - 6(1) - 3}$$ $$L = \sqrt{25 + 1 + 20 - 6 - 3}$$ $$L = \sqrt{37}$$Normal to a Circle
A normal is a line perpendicular to the tangent at the point of contact.
Key Property: The normal to a circle always passes through the center.
Visual Description: If a wheel is on the road, the normal is the spoke from the point touching the road to the center of the wheel.
Equation of Normal at Point $(x_1, y_1)$
For circle $x^2 + y^2 = r^2$:
$$\boxed{\frac{x}{x_1} = \frac{y}{y_1}}$$or
$$\boxed{xy_1 - yx_1 = 0}$$For general circle with center $(h, k)$:
The normal passes through $(x_1, y_1)$ and $(h, k)$:
$$\boxed{\frac{y - y_1}{x - x_1} = \frac{y_1 - k}{x_1 - h}}$$Shortcut: Since normal passes through center, use two-point form of straight line!
Chord Properties
1. Equation of Chord
Chord with midpoint $(x_1, y_1)$ for circle $x^2 + y^2 = r^2$:
$$\boxed{xx_1 + yy_1 = x_1^2 + y_1^2}$$(This is called chord of contact formula when applied differently)
2. Length of Chord
For chord at perpendicular distance $d$ from center of circle with radius $r$:
$$\boxed{L = 2\sqrt{r^2 - d^2}}$$Visual Description: This is just the Pythagorean theorem! Half the chord, the perpendicular from center, and the radius form a right triangle.
Find length of chord of circle $x^2 + y^2 = 25$ at distance $3$ from center.
Solution:
Radius $r = 5$, distance $d = 3$
$$L = 2\sqrt{5^2 - 3^2} = 2\sqrt{25 - 9} = 2\sqrt{16} = 2 \times 4 = 8$$3. Chord of Contact
From external point $(x_1, y_1)$, draw tangents to circle. The chord joining the two points of contact is called chord of contact.
Equation for circle $x^2 + y^2 = r^2$:
$$\boxed{xx_1 + yy_1 = r^2}$$For general circle $x^2 + y^2 + 2gx + 2fy + c = 0$:
$$\boxed{xx_1 + yy_1 + g(x+x_1) + f(y+y_1) + c = 0}$$Visual Description: Stand outside a circular pond and throw two ropes to touch it tangentially. The chord of contact is the line joining those two touching points.
Position of Point Relative to Circle
For circle $x^2 + y^2 + 2gx + 2fy + c = 0$ and point $(x_1, y_1)$:
Let $S_1 = x_1^2 + y_1^2 + 2gx_1 + 2fy_1 + c$
- If $S_1 > 0$ → Point is outside the circle
- If $S_1 = 0$ → Point is on the circle
- If $S_1 < 0$ → Point is inside the circle
Why? Compare $S_1$ with radius squared!
Family of Circles
1. Concentric Circles
Circles with same center but different radii:
If $x^2 + y^2 + 2gx + 2fy + c = 0$ is given, family of concentric circles:
$$\boxed{x^2 + y^2 + 2gx + 2fy + k = 0}$$(Same $g$ and $f$, vary $k$)
2. Circles Through Intersection
Circles passing through intersection points of circles $S_1 = 0$ and $S_2 = 0$:
$$\boxed{S_1 + \lambda S_2 = 0}$$or
$$\boxed{x^2 + y^2 + 2gx + 2fy + c + \lambda(x^2 + y^2 + 2g'x + 2f'y + c') = 0}$$Special Case: If $\lambda = -1$, we get the radical axis (common chord):
$$\boxed{S_1 - S_2 = 0}$$This is a straight line!
Find radical axis of circles $x^2 + y^2 - 4x - 6y + 9 = 0$ and $x^2 + y^2 - 2x - 4y + 3 = 0$.
Solution:
$S_1 - S_2 = 0$:
$$(x^2 + y^2 - 4x - 6y + 9) - (x^2 + y^2 - 2x - 4y + 3) = 0$$ $$-4x - 6y + 9 + 2x + 4y - 3 = 0$$ $$\boxed{-2x - 2y + 6 = 0 \quad \text{or} \quad x + y = 3}$$Memory Tricks & Patterns
Pattern 1: Tangent Formula
Transform the circle equation by replacing:
| In Circle Equation | In Tangent at $(x_1, y_1)$ |
|---|---|
| $x^2$ | $xx_1$ |
| $y^2$ | $yy_1$ |
| $x$ | $\frac{x + x_1}{2}$ |
| $y$ | $\frac{y + y_1}{2}$ |
Example: $x^2 + y^2 - 6x + 4y - 12 = 0$ at point $(1, 1)$
Tangent: $x(1) + y(1) - 3(x+1) + 2(y+1) - 12 = 0$
Simplify: $x + y - 3x - 3 + 2y + 2 - 12 = 0$
$$\boxed{-2x + 3y - 13 = 0}$$Pattern 2: Normal Always Passes Through Center
“Normal is NO-PROBLEM - it always goes through center!”
To write normal equation:
- Find center of circle
- Use two-point form with given point and center
Pattern 3: Chord Length Formula
“Right triangle: Half-chord, Perpendicular, Radius”
$$r^2 = \left(\frac{L}{2}\right)^2 + d^2$$where $L$ = chord length, $d$ = perpendicular distance, $r$ = radius
Common Mistakes to Avoid
Tangent at a point $(x_1, y_1)$ on the circle:
$$xx_1 + yy_1 = r^2$$Chord of contact from external point $(x_1, y_1)$:
$$xx_1 + yy_1 = r^2$$(same formula!)
Difference:
- For tangent: $(x_1, y_1)$ is on the circle
- For chord of contact: $(x_1, y_1)$ is outside the circle
The formula looks same, but context differs!
Wrong: $r = \sqrt{g + f - c}$
Correct: $r = \sqrt{g^2 + f^2 - c}$
Remember: It’s $g$ squared plus $f$ squared, not just $g + f$!
In $x^2 + y^2 + 2gx + 2fy + c = 0$:
Wrong: Center = $(g, f)$
Correct: Center = $(-g, -f)$
Memory: “Go and Find with MINUS signs!”
Students often confuse “tangent at a point” with “tangent from a point”:
- At a point on circle: Use tangent formula directly
- From external point: Two tangents exist, use $y = mx \pm r\sqrt{1+m^2}$ or chord of contact
Check: Is the point on the circle or outside? This determines which method to use.
Practice Problems
Level 1: Foundation (NCERT Style)
Question: Find center and radius of circle $x^2 + y^2 - 6x + 4y - 12 = 0$.
Solution:
General form: $x^2 + y^2 + 2gx + 2fy + c = 0$
Comparing: $2g = -6$ → $g = -3$
$2f = 4$ → $f = 2$
$c = -12$
Center: $(-g, -f) = (3, -2)$
Radius: $r = \sqrt{g^2 + f^2 - c} = \sqrt{9 + 4 + 12} = \sqrt{25} = 5$
$$\boxed{\text{Center: }(3, -2), \quad \text{Radius: } 5}$$Question: Find equation of circle with diameter endpoints $(2, 3)$ and $(4, 5)$.
Solution:
Using diameter form:
$$(x - 2)(x - 4) + (y - 3)(y - 5) = 0$$ $$x^2 - 6x + 8 + y^2 - 8y + 15 = 0$$ $$\boxed{x^2 + y^2 - 6x - 8y + 23 = 0}$$Verification: Center = midpoint = $\left(\frac{2+4}{2}, \frac{3+5}{2}\right) = (3, 4)$
From general form: $g = -3$, $f = -4$ → Center $= (3, 4)$ ✓
Question: Find equation of tangent to $x^2 + y^2 = 25$ at point $(3, 4)$.
Solution:
Using tangent formula $xx_1 + yy_1 = r^2$:
$$x(3) + y(4) = 25$$ $$\boxed{3x + 4y = 25}$$Verification: Check if $(3,4)$ satisfies circle: $9 + 16 = 25$ ✓
Level 2: JEE Main Type
Question: Determine if point $(2, 3)$ lies inside, on, or outside the circle $x^2 + y^2 - 4x - 6y + 9 = 0$.
Solution:
Calculate $S_1 = x_1^2 + y_1^2 + 2gx_1 + 2fy_1 + c$
Here: $2g = -4$, $2f = -6$, $c = 9$
$$S_1 = 2^2 + 3^2 - 4(2) - 6(3) + 9$$ $$S_1 = 4 + 9 - 8 - 18 + 9 = -4$$Since $S_1 < 0$:
$$\boxed{\text{Point lies INSIDE the circle}}$$Question: Find equation of circle passing through $(0, 0)$, $(a, 0)$, and $(0, b)$.
Solution:
Assume $x^2 + y^2 + 2gx + 2fy + c = 0$
At $(0, 0)$: $c = 0$
At $(a, 0)$: $a^2 + 2ga = 0$ → $a(a + 2g) = 0$ → $g = -\frac{a}{2}$
At $(0, b)$: $b^2 + 2fb = 0$ → $b(b + 2f) = 0$ → $f = -\frac{b}{2}$
Equation: $x^2 + y^2 - ax - by = 0$
Multiply by appropriate factor to clear fractions:
$$\boxed{x^2 + y^2 - ax - by = 0}$$Note: This can also be solved using diameter form since $(0,0)$ and center make special geometry.
Question: Find equation of chord of contact of tangents drawn from $(5, 4)$ to circle $x^2 + y^2 = 16$.
Solution:
Using chord of contact formula $xx_1 + yy_1 = r^2$:
$$x(5) + y(4) = 16$$ $$\boxed{5x + 4y = 16}$$Note: Point $(5, 4)$ is outside since $25 + 16 = 41 > 16$.
Level 3: JEE Advanced Type
Question: Find equation of circle passing through intersection of circles $x^2 + y^2 = 4$ and $x^2 + y^2 - 2x - 4y + 4 = 0$, and also passing through origin.
Solution:
Family through intersection:
$$S_1 + \lambda S_2 = 0$$ $$(x^2 + y^2 - 4) + \lambda(x^2 + y^2 - 2x - 4y + 4) = 0$$Since it passes through origin $(0, 0)$:
$$-4 + \lambda(4) = 0$$ $$\lambda = 1$$Circle equation:
$$(x^2 + y^2 - 4) + (x^2 + y^2 - 2x - 4y + 4) = 0$$ $$2x^2 + 2y^2 - 2x - 4y = 0$$ $$\boxed{x^2 + y^2 - x - 2y = 0}$$Question: Find $k$ if circles $x^2 + y^2 + 2x + 3y + k = 0$ and $x^2 + y^2 + 4x + y + 1 = 0$ are orthogonal.
Solution:
Two circles are orthogonal if they intersect at right angles.
Condition: $2g_1g_2 + 2f_1f_2 = c_1 + c_2$
Circle 1: $g_1 = 1$, $f_1 = \frac{3}{2}$, $c_1 = k$
Circle 2: $g_2 = 2$, $f_2 = \frac{1}{2}$, $c_2 = 1$
$$2(1)(2) + 2\left(\frac{3}{2}\right)\left(\frac{1}{2}\right) = k + 1$$ $$4 + \frac{3}{2} = k + 1$$ $$\frac{11}{2} = k + 1$$ $$\boxed{k = \frac{9}{2}}$$Question: Find the locus of point of intersection of perpendicular tangents to circle $x^2 + y^2 = a^2$.
Solution:
Let $(h, k)$ be the point from which perpendicular tangents are drawn.
Tangents from $(h, k)$ to circle $x^2 + y^2 = a^2$ have equations:
Using slope form, if slopes are $m_1$ and $m_2$:
$$k = m_1 h \pm a\sqrt{1 + m_1^2}$$ $$k = m_2 h \pm a\sqrt{1 + m_2^2}$$For perpendicular tangents: $m_1 m_2 = -1$
After algebraic manipulation (using condition that both equations are satisfied):
The locus is a circle with double the radius:
$$\boxed{x^2 + y^2 = 2a^2}$$This is called the Director Circle.
Replace $(h,k)$ with $(x,y)$ for final answer.
Quick Revision Box
| Situation | Formula/Approach |
|---|---|
| Standard form | $(x-h)^2 + (y-k)^2 = r^2$ |
| General form | $x^2 + y^2 + 2gx + 2fy + c = 0$ |
| Center from general | $(-g, -f)$ |
| Radius from general | $\sqrt{g^2 + f^2 - c}$ |
| Tangent at $(x_1,y_1)$ | $xx_1 + yy_1 = r^2$ |
| Normal at point | Passes through center |
| Chord of contact | $xx_1 + yy_1 = r^2$ (external point) |
| Length of tangent | $\sqrt{x_1^2 + y_1^2 + 2gx_1 + 2fy_1 + c}$ |
| Chord length | $2\sqrt{r^2 - d^2}$ |
| Radical axis | $S_1 - S_2 = 0$ |
| Parametric form | $x = h + r\cos\theta$, $y = k + r\sin\theta$ |
Cross-Links to Related Topics
Prerequisites:
- Straight Lines - For tangent and normal equations
- Trigonometric Identities - For parametric equations
Related Topics:
- Parabola - Next conic section
- Ellipse - Circle is special case ($a = b$)
- Hyperbola - Another conic section
- General Conics - Unified conic approach
- Tangent and Normal to Conics - Tangent techniques
- Chord of Contact - Chord properties
- 3D Coordinates - Spheres are 3D circles
Applications in Physics:
- Circular Motion - Motion on circular paths
- Gravitation - Planetary orbits
Applications in Math:
- Complex Numbers - Argand Diagram - Circle in complex plane
JEE Exam Strategy
Weightage: 4-5 questions in JEE Main, 2-3 in Advanced
Most Frequently Asked:
- Equation of tangent/normal (appears almost every year)
- Chord of contact and length of tangent
- Family of circles through intersection
- Position of point (inside/outside/on circle)
- Orthogonal circles
Time-Saving Tricks:
- Tangent formula: Just replace $x^2$ with $xx_1$, $y^2$ with $yy_1$
- Normal: Always passes through center - use two-point form
- Check position quickly: Substitute point in $x^2 + y^2 + 2gx + 2fy + c$ and check sign
Common Trap Options:
- Giving center as $(g, f)$ instead of $(-g, -f)$
- Missing the square in radius formula: $\sqrt{g^2 + f^2 - c}$
- Confusing tangent at point vs chord of contact from external point
- Wrong sign in diameter form
Pattern Recognition:
- If problem gives “tangents are perpendicular” → Use director circle
- If “two circles intersect orthogonally” → Use $2g_1g_2 + 2f_1f_2 = c_1 + c_2$
- If “find equation through three points” → Use family of circles (faster than solving three equations)
Teacher’s Summary
- Master the tangent formula - Replace squares with products: $x^2 \to xx_1$, $y^2 \to yy_1$
- Remember center has negative signs - In $x^2 + y^2 + 2gx + 2fy + c = 0$, center is $(-g, -f)$
- Normal is easy - It always passes through center, so use two-point form
- Family of circles - Use $S_1 + \lambda S_2 = 0$ to avoid finding intersection points explicitly
- Position check - Substitute point in circle equation; positive = outside, zero = on, negative = inside
“The circle is the first conic section - master it well, because parabolas, ellipses, and hyperbolas build on these same ideas!”
Next Step: Ready to explore Parabola - where the satellite dish shape comes from!