high school Equations of a Circle

Circle Equations

A circle is the path of every point lying the same distance (the radius) from one center. With center (a, b) and radius r it becomes (x − a)² + (y − b)² = r², straight from the distance formula. An expanded general form turns readable again once you complete the square to recover the center and radius. Drag the center and radius and the equation and the circle change together.

Equation of a Circle — Center & Radius

A circle is the set of points equidistant from a center. Knowing the center (a, b) and radius r determines the equation.

Center x2

Center y1

Radius r3

💡 Deriving the Equation

①Distance from (x, y) on circle to center (a, b) = r

②√((x-a)² + (y-b)²) = r

③Square both sides → (x-a)² + (y-b)² = r²

Standard Form & General Form

Standard Form

(x - a)² + (y - b)² = r²

Center (a, b), radius r

General Form

x² + y² + Dx + Ey + F = 0

Expanded: D=-2a, E=-2b, F=a²+b²-r²

📐 General → Standard

①Group x: (x + D/2)²

②Group y: (y + E/2)²

③r² = (D/2)² + (E/2)² - F

④Need r² > 0 for the circle to exist

⑤The line through two circles' intersections is f₁ − f₂ = 0. Tangent formulas are in the boxes below

Circle vs Line Relations

The distance d from the center to the line, compared with radius r, determines the relation.

Distance d2

Circle–Line Position

d < r: two points, d = r: tangent, d > r: no intersection

d = distance from center to the line

Tangent Lines

Tangent at a Point on the Circle

x₁·x + y₁·y = r²

For a point (x₁,y₁) on x²+y²=r²

Tangent with Slope m

y = mx ± r√(1 + m²)

Line of slope m tangent to the origin-centered circle

Work It Out

Example 1

Find the equation of the circle with center (2, −1) and radius 3.

Substitute a = 2, b = −1, r = 3 into the standard form (x − a)² + (y − b)² = r². Since b is negative, it becomes (y + 1).

(x − 2)² + (y + 1)² = 3²

Evaluate the right side.

(x − 2)² + (y + 1)² = 9

▸ (x − 2)² + (y + 1)² = 9

The center coordinates enter the equation with flipped signs.

Example 2

Find the center and radius of the circle x² + y² − 6x + 4y − 12 = 0.

Complete the square in x and in y.

(x − 3)² − 9 + (y + 2)² − 4 − 12 = 0

Move the constants to the right to reach standard form.

(x − 3)² + (y + 2)² = 25, center (3, −2), radius 5

▸ Center (3, −2), radius 5

Turn the general form into standard form by completing the square.

Wrap-up

Standard Form

(x - a)² + (y - b)² = r²

Center & radius visible

Tangency Test

distance d from center to line = r

Tangency condition

Grade-10 school exam type

From the point (5, 5), find the length of the tangent to the circle (x − 1)² + (y − 2)² = 9.

①3

②4

③5

④6

⑤7

▸ ② 4

First find the distance d from the external point to the center.

d = √((5 − 1)² + (5 − 2)²) = √25 = 5

The tangent length is √(d² − r²).

√(d² − r²) = √(25 − 9) = 4

🎯 Exam Points

①General → standard via completing the square

②Verify r² > 0 (circle exists)

③Circle vs line: also use discriminant D from substitution

④Tangent: point-on-circle vs external-point cases

⑤Line through intersection of two circles: f₁ - f₂ = 0

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Plane Coordinates & Line Equations

Geometric Transformations

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