Equation of a circle

LVL: FREE

MODULE: Coordinate Geometry and Vectors

✖️ 1. Deriving the standard circle equation from the distance formula

📐 Deriving the Standard Circle Equation

  • A circle is all points at a fixed distance from a center point.
  • Use the distance formula between any point (x,y)(x,y) and center (h,k)(h,k).
  • Set distance equal to radius: (x−h)2+(y−k)2=r\sqrt{(x-h)^2 + (y-k)^2} = r.
  • Square both sides to eliminate the square root.
  • Final form: (x−h)2+(y−k)2=r2(x-h)^2 + (y-k)^2 = r^2.

Example: Center at (3,−2)(3,-2) with radius 55 gives (x−3)2+(y+2)2=25(x-3)^2 + (y+2)^2 = 25.

💡 Think: Distance formula squared = circle equation!

1. Deriving the standard circle equation from the distance formula

Deriving the Standard Circle Equation

A circle is the set of all points in a plane equidistant from a fixed point called the center. The standard equation (x−h)2+(y−k)2=r2(x-h)^2 + (y-k)^2 = r^2 emerges directly from applying the distance formula to any point (x,y)(x,y) on the circle and the center (h,k)(h,k).

Intuition: Every point on the circle maintains exactly distance rr from (h,k)(h,k), so the distance formula (x−h)2+(y−k)2=r\sqrt{(x-h)^2 + (y-k)^2} = r becomes the circle equation when squared.

Core Derivation Steps:

  • Start with distance formula: (x−h)2+(y−k)2=r\sqrt{(x-h)^2 + (y-k)^2} = r
  • Square both sides to eliminate the radical: (x−h)2+(y−k)2=r2(x-h)^2 + (y-k)^2 = r^2
  • The center (h,k)(h,k) can be any point; r>0r > 0 is required for a valid circle
  • When h=0h=0 and k=0k=0, the equation simplifies to x2+y2=r2x^2 + y^2 = r^2 (circle centered at origin)

Consequence: This form immediately reveals geometric properties and is the foundation for all circle analysis.

Example: A circle centered at (3,−2)(3,-2) with radius 55 has equation (x−3)2+(y+2)2=25(x-3)^2 + (y+2)^2 = 25.

Progress0 / 3
Level 2
MOD: TRANSLATE

A line has a slope of m=4m = 4 and passes through the point (2,9)(2, 9). Using the point-slope form, find the value of the yy-intercept bb when the equation is converted to slope-intercept form.

Deep reasoning
Ultra

✖️ 2. Extracting the center and radius from standard form

🎯 Reading Center and Radius from Standard Form

  • Standard form is (x−h)2+(y−k)2=r2(x-h)^2 + (y-k)^2 = r^2.
  • The center is (h,k)(h,k) — use the opposite sign of what appears.
  • The radius is r=r2r = \sqrt{r^2} — take the square root of the right side.
  • If you see (x+4)(x+4), then h=−4h = -4 (flip the sign).
  • If you see (y−7)(y-7), then k=7k = 7 (flip the sign).

Example: (x+1)2+(y−5)2=49(x+1)^2 + (y-5)^2 = 49 has center (−1,5)(-1, 5) and radius 77.

💡 Flip signs for center, square root for radius!

2. Extracting the center and radius from standard form

Extracting Center and Radius from Standard Form

The standard form (x−h)2+(y−k)2=r2(x-h)^2 + (y-k)^2 = r^2 encodes the circle's center and radius through its algebraic structure. Reading these values requires careful attention to signs within the squared binomials.

Intuition: The values subtracted inside each squared term give the center coordinates directly, while the right side gives the squared radius.

Core Extraction Rules:

  • Center hh-coordinate: The value subtracted from xx (if (x−h)(x-h), then hh is positive; if (x+h)(x+h), then hh is negative)
  • Center kk-coordinate: The value subtracted from yy (same sign logic applies)
  • Radius: r=right sider = \sqrt{\text{right side}} (always take the positive square root)
  • The right side must be positive for a real circle; if zero, the circle degenerates to a point

Consequence: Misreading signs is the most common error; (x+3)2(x+3)^2 means h=−3h=-3, not h=3h=3.

Example: From (x+4)2+(y−1)2=49(x+4)^2 + (y-1)^2 = 49, the center is (−4,1)(-4,1) and radius is r=49=7r=\sqrt{49}=7.

Progress0 / 3
Level 2
STRC: TRANSFORM

Convert the equation y=−5x+7y = -5x + 7 into standard form Ax+By=CAx + By = C.

Deep reasoning
Ultra

✖️ 3. Converting general form to standard form via completing the square

🔄 Converting General Form to Standard Form

  • General form: x2+y2+Dx+Ey+F=0x^2 + y^2 + Dx + Ey + F = 0.
  • Group xx terms together and yy terms together.
  • Complete the square for both groups separately.
  • Add the same values to both sides to keep equality.
  • Rewrite as (x−h)2+(y−k)2=r2(x-h)^2 + (y-k)^2 = r^2.

Example: x2+y2+6x−4y−3=0x^2 + y^2 + 6x - 4y - 3 = 0 becomes (x+3)2+(y−2)2=16(x+3)^2 + (y-2)^2 = 16 after completing the square.

💡 Complete the square twice, balance both sides!

3. Converting general form to standard form via completing the square

Converting General Form to Standard Form

The general form x2+y2+Dx+Ey+F=0x^2 + y^2 + Dx + Ey + F = 0 obscures the center and radius. Converting to standard form requires completing the square separately for xx and yy terms.

Intuition: Rearrange terms into groups, then add strategic constants to form perfect square trinomials, revealing (x−h)2(x-h)^2 and (y−k)2(y-k)^2 patterns.

Core Conversion Steps:

  • Group xx terms and yy terms: (x2+Dx)+(y2+Ey)=−F(x^2 + Dx) + (y^2 + Ey) = -F
  • Complete the square for xx: add (D/2)2(D/2)^2 to both sides
  • Complete the square for yy: add (E/2)2(E/2)^2 to both sides
  • Rewrite as (x+D/2)2+(y+E/2)2=(D/2)2+(E/2)2−F(x + D/2)^2 + (y + E/2)^2 = (D/2)^2 + (E/2)^2 - F
  • Validity check: Right side must be positive; if negative, no real circle exists

Consequence: The center is (−D/2,−E/2)(-D/2, -E/2) and radius is r=(D/2)2+(E/2)2−Fr = \sqrt{(D/2)^2 + (E/2)^2 - F}.

Example: For x2+y2−6x+4y−3=0x^2 + y^2 - 6x + 4y - 3 = 0, completing the square yields (x−3)2+(y+2)2=16(x-3)^2 + (y+2)^2 = 16, so center (3,−2)(3,-2) and r=4r=4.

Progress0 / 3
Level 2
MOD: TRANSLATE

A line is perpendicular to y=4x−5y = 4x - 5. What is its slope?

Deep reasoning
Ultra

✖️ 4. Applications in telecommunications and seismology

📡 Real-World Applications of Circle Equations

  • Telecommunications: Cell towers broadcast in circular coverage areas.
  • The tower location is the center (h,k)(h,k).
  • The broadcast range is the radius rr.
  • Seismology: Earthquake epicenters create circular wave patterns.
  • Scientists use circle equations to triangulate the epicenter location.

Example: A tower at (10,20)(10, 20) with 15 km range is (x−10)2+(y−20)2=225(x-10)^2 + (y-20)^2 = 225.

💡 Center = source location, radius = reach distance!

4. Applications in telecommunications and seismology

Applications: Coverage Areas and Epicenter Radii

Circle equations model real-world phenomena where distance from a central point determines a boundary. Telecommunications and seismology rely on these geometric models for planning and analysis.

Intuition: Any scenario involving "all points within distance rr from location (h,k)(h,k)" translates directly to a circle equation.

Core Applications:

  • Broadcast towers: A tower at (h,k)(h,k) with range rr kilometers serves all points satisfying (x−h)2+(y−k)2≤r2(x-h)^2 + (y-k)^2 \leq r^2
  • Seismic epicenters: An earthquake epicenter at (h,k)(h,k) with intensity radius rr affects regions within (x−h)2+(y−k)2=r2(x-h)^2 + (y-k)^2 = r^2
  • Coverage optimization: Multiple towers require solving systems of circle equations to eliminate dead zones
  • Coordinates typically use projected map systems (kilometers or miles from a reference point)

Consequence: Engineers use these equations to calculate infrastructure placement and predict impact zones with precision.

Example: A cell tower at (10,15)(10,15) with 8-kilometer range covers all points in (x−10)2+(y−15)2≤64(x-10)^2 + (y-15)^2 \leq 64.

Progress0 / 3
Level 2
MOD: TRANSLATE

A delivery van is purchased for 35000 dollars. Its value depreciates linearly by 3000 dollars each year. What is the value of the van, in dollars, after 6 years?

Deep reasoning
Ultra

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