Derive the equation of a circle from the Pythagorean Theorem and complete the square to find a center and radius

Learning goal: Derive the equation of a circle of given center and radius using the Pythagorean Theorem; complete the square to find the center and radius of a circle given by an equation; and verify algebraically whether a specific point lies on a given circle.

A circle is a distance condition. A circle with center C(h, k) and radius r is the set of points whose distance from C is exactly r. Put a point P(x, y) on the circle and draw the horizontal segment from C to (x, k) and the vertical segment from (x, k) to P. Those segments have lengths |x − h| and |y − k| and meet at a right angle, and the hypotenuse CP has length r. The Pythagorean Theorem gives (x − h)² + (y − k)² = r². (The absolute values disappear because squaring a number and squaring its absolute value give the same result.) The equation holds for every point on the circle, and conversely any point satisfying it is at distance r from C, so it is on the circle. That two-way statement makes (x − h)² + (y − k)² = r² the equation of the circle; it is called the standard form. If the center is the origin it reduces to x² + y² = r².

Derive, then expand. The circle with center (3, −2) and radius 5 has standard form (x − 3)² + (y − (−2))² = 5², that is (x − 3)² + (y + 2)² = 25. Watch the signs: the center's coordinates appear with opposite signs inside the parentheses. Expanding, x² − 6x + 9 + y² + 4y + 4 = 25, so x² + y² − 6x + 4y − 12 = 0. This is the general form: the squared terms have equal coefficients, there is no xy term, and the center and radius are no longer visible. The next skill recovers them.

Complete the square to find center and radius. Start from x² + y² − 6x + 4y − 12 = 0. Group the x terms and the y terms and move the constant across: (x² − 6x) + (y² + 4y) = 12. To make x² − 6x a perfect square, add (−6/2)² = 9; to make y² + 4y a perfect square, add (4/2)² = 4. Whatever is added to the left side must be added to the right: (x² − 6x + 9) + (y² + 4y + 4) = 12 + 9 + 4, so (x − 3)² + (y + 2)² = 25. The center is (3, −2) and the radius is √25 = 5, as it should be. The step students skip is adding the same numbers to both sides; forgetting to add 9 and 4 on the right gives r² = 12 and a wrong radius.

A worked model inside the lesson. Find the center and radius of the circle 2x² + 2y² − 8x + 12y − 6 = 0. First divide by 2 so the squared terms have coefficient 1: x² + y² − 4x + 6y − 3 = 0. Group and move: (x² − 4x) + (y² + 6y) = 3. Add (−4/2)² = 4 and (6/2)² = 9 to both sides: (x − 2)² + (y + 3)² = 3 + 4 + 9 = 16. Center (2, −3), radius 4. Check with one point: (6, −3) is 4 units to the right of the center, so it should satisfy the original equation: 2·36 + 2·9 − 48 − 36 − 6 = 72 + 18 − 48 − 36 − 6 = 0. It does.

When the equation is not a circle. Completing the square on x² + y² − 2x + 4y + 9 = 0 gives (x − 1)² + (y + 2)² = −9 + 1 + 4 = −4. A sum of two squares cannot be negative, so no point satisfies the equation; the graph is empty. If the right side had come out 0, the graph would be the single point (1, −2). A genuine circle needs r² > 0. State this condition whenever you report a center and radius.

Is a point on the circle? Substitute the point into the standard form and compare with r². For the circle (x − 3)² + (y + 2)² = 25: the point (7, 1) gives (4)² + (3)² = 16 + 9 = 25, so it is on the circle. The point (6, 2) gives 9 + 16 = 25, also on the circle. The point (5, 3) gives 4 + 25 = 29 > 25, so it is outside the circle (farther than 5 from the center); a point giving less than 25 would be inside. This is the algebraic way to verify a geometric statement about a circle: "P is on the circle" becomes "P satisfies the equation."

From other information to the equation. If the endpoints of a diameter are (1, 4) and (7, −4), the center is their midpoint, ((1 + 7)/2, (4 + (−4))/2) = (4, 0), and the radius is half the diameter: ½√((7 − 1)² + (−4 − 4)²) = ½√(36 + 64) = ½·10 = 5. The equation is (x − 4)² + y² = 25. If instead the center (−1, 2) and one point on the circle (2, 6) are given, r² is the squared distance between them: 3² + 4² = 25, so (x + 1)² + (y − 2)² = 25; there is no need to take the square root and square again.

The misconception: the equation gives the center's coordinates directly. In (x − 3)² + (y + 2)² = 25 the center is (3, −2), not (−3, 2) or (3, 2). Substituting x = 3 and y = −2 makes both squared terms zero, which is exactly what should happen at the center, distance 0 from itself. Reading the center by asking "what values make each parenthesis zero?" avoids the sign error. A related slip is reporting r = 25; the right side is r², so the radius is 5.

Worked derivation: Circle with center (−5, 1) and radius 5. Standard form: (x + 5)² + (y − 1)² = 25. Expand: x² + 10x + 25 + y² − 2y + 1 = 25, so the general form is x² + y² + 10x − 2y + 1 = 0. Reverse it: (x² + 10x) + (y² − 2y) = −1; add 25 and 1 to both sides: (x + 5)² + (y − 1)² = −1 + 25 + 1 = 25. Center (−5, 1), radius 5, as we started.

Worked completing the square with fractions: x² + y² + 3x − 5y + 2 = 0. Group: (x² + 3x) + (y² − 5y) = −2. Add (3/2)² = 9/4 and (5/2)² = 25/4 to both sides: (x + 3/2)² + (y − 5/2)² = −2 + 9/4 + 25/4 = −8/4 + 34/4 = 26/4 = 13/2. Center (−3/2, 5/2), radius √(13/2) = √26/2 ≈ 2.55. Where is the origin? Substituting (0, 0): (3/2)² + (−5/2)² = 9/4 + 25/4 = 34/4 = 8.5, and 8.5 > 6.5 = r², so the origin is outside the circle, at distance √8.5 ≈ 2.92 from the center, more than the radius 2.55.

Worked point test: Does (−1, 4) lie on the circle x² + y² − 6x + 4y − 12 = 0? The standard form is (x − 3)² + (y + 2)² = 25. Substituting: (−4)² + 6² = 16 + 36 = 52 ≠ 25. The point is not on the circle; since 52 > 25 it is outside, at distance √52 ≈ 7.21 from the center. The point (0, 2) gives 9 + 16 = 25, so (0, 2) is on the circle.

Worked tangent line from the equation: Find the tangent to (x − 3)² + (y + 2)² = 25 at the point (7, 1). The radius from (3, −2) to (7, 1) has slope 3/4, so the tangent, perpendicular to the radius, has slope −4/3 and passes through (7, 1): y − 1 = −(4/3)(x − 7), or 4x + 3y = 31. Check that (7, 1) satisfies it: 28 + 3 = 31.

Two-way record: Make a table with columns center, radius, standard form, general form. Fill three rows starting from a center and radius of your choice (one with a negative coordinate, one with a fractional radius such as 3/2) and three rows starting from a general-form equation (one whose squared terms have coefficient 3, one that turns out to be a single point). Show the completing-the-square steps beneath the table for the general-form rows.

Independent performance: Derive the equation of a circle from the Pythagorean Theorem in your own words with a labeled diagram, using the general center (h, k) and radius r; then write the circle through the three points (0, 0), (8, 0) and (0, 6) by finding its center and radius, expand to general form, and complete the square to recover the center. Finally test three points of your choosing, one on, one inside and one outside, and say which is which. Success criteria: the derivation names the right angle and states both directions (on the circle implies the equation, and the equation implies on the circle); the completing-the-square work shows the same quantities added to both sides; and each point test compares the computed value with r² and interprets the comparison.