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Grade 10 / High 1 (age 15-16)

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 here to see 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.

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💡 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
Circle vs Line Relations

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

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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.
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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
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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.
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Complete the square in x and in y.
(x − 3)² − 9 + (y + 2)² − 4 − 12 = 0
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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.
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② 4
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First find the distance d from the external point to the center.
d = √((5 − 1)² + (5 − 2)²) = √25 = 5
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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 & Lines
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Geometric Transformations
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