Circle equation — standard form (x−h)² + (y−k)² = r²: Exercises

Before we begin, two tools appear again and again — let us define both on this page so no step is a mystery.

Definition Two tools we use everywhere

The Distance Formula. The straight-line distance between two points $(x_{1}, y_{1})$ and $(x_{2}, y_{2})$ is $d = (x_{2} - x_{1})^{2} + (y_{2} - y_{1})^{2} .$ In words: "how far apart horizontally" squared plus "how far apart vertically" squared, then square-rooted. It is just the Pythagorean Theorem applied to the right triangle whose legs are the horizontal gap $x_{2} - x_{1}$ and the vertical gap $y_{2} - y_{1}$ , with $d$ as the slanted hypotenuse.

Completing the Square. Any expression $x^{2} + b x$ can be turned into a perfect square by adding one specific number: take half of $b$ , then square it. That is, $x^{2} + b x + (\frac{b}{2})^{2} = (x + \frac{b}{2})^{2}$ . Why half of $b$ ? Because $(x + a)^{2} = x^{2} + 2 a x + a^{2}$ : the middle term is $2 a$ , so to match $b x$ we need $a = \frac{b}{2}$ , and the constant that finishes the pattern is $a^{2} = (\frac{b}{2})^{2}$ .

Now one picture to fix the meaning of every letter we will use.

Figure — Circle equation — standard form (x−h)² + (y−k)² = r²

Figure s01 — a lavender circle with centre $(h, k)$ marked as a coral dot. A mint segment runs from the centre out to a point $(x, y)$ on the rim: that segment is the radius $r$ . Two butter-coloured legs form a right triangle under it — a horizontal leg of length $x - h$ and a vertical leg of length $y - k$ — so Pythagoras gives $(x - h)^{2} + (y - k)^{2} = r^{2}$ .

Level 1 — Recognition

Just read the pieces off the page. No algebra beyond a square root.

Recall Solution 1.1

WHAT we do: match the equation to $(x - h)^{2} + (y - k)^{2} = r^{2}$ and read letters off.

$(x - 4)^{2} \Rightarrow h = 4$ (minus in bracket → positive centre).
$(y - 1)^{2} \Rightarrow k = 1$ .
Right side is $r^{2} = 9$ , so $r = 9 = 3$ .

WHY take the square root: the right side of the standard form is $r^{2}$ , not $r$ — it stores the radius already squared (because the derivation squared both sides of the distance equation to kill the root). To recover the actual rope-length $r$ we must undo that squaring, and the operation that undoes squaring is the square root. So $r = 9 = 3$ .

Answer: centre $(4, 1)$ , radius $3$ .

Recall Solution 1.2

WHY the sign flip: the standard form has minus signs, so we rewrite the plus signs to see the truth: $(x + 2)^{2} = (x - (- 2))^{2}, (y + 6)^{2} = (y - (- 6))^{2} .$

$h = - 2$ , $k = - 6$ .
$r^{2} = 49 \Rightarrow r = 49 = 7$ .

Answer: centre $(- 2, - 6)$ , radius $7$ .

Recall Solution 1.3

Centre $(0, 0)$ means $h = 0$ and $k = 0$ , so substitute these into the standard form: $(x - 0)^{2} + (y - 0)^{2} = 6^{2} \Rightarrow x^{2} + y^{2} = 36.$ WHY the form collapses: subtracting zero changes nothing — $x - 0$ is just $x$ and $y - 0$ is just $y$ . The centre's coordinates were the only things doing the shifting in $(x - h)$ and $(y - k)$ ; with both set to zero there is nothing to shift, so the circle sits symmetrically around the origin and the brackets simplify straight to $x$ and $y$ . Answer: $x^{2} + y^{2} = 36$ .

Level 2 — Application

Plug given data into the formula and turn a handle.

Recall Solution 2.1

WHAT: substitute $h = - 3$ , $k = 5$ into the form and square the radius. $(x - (- 3))^{2} + (y - 5)^{2} = (25)^{2} .$ Simplify: $x - (- 3) = x + 3$ , and $(25)^{2} = 2^{2} \cdot (5)^{2} = 4 \cdot 5 = 20$ . $(x + 3)^{2} + (y - 5)^{2} = 20$ WHY square the radius: the form comes from squaring the distance equation, so the right side is $r^{2}$ , not $r$ .

Recall Solution 2.2

WHY this test works: the left side $(x - 3)^{2} + (y + 1)^{2}$ is the squared distance from the point to the centre $(3, - 1)$ . The circle is exactly the set of points whose squared distance equals $r^{2} = 25$ . So if we plug the point in and the left side comes out to $25$ , the point is precisely $r$ away from the centre — on the rim. If it were smaller than $25$ the point would sit inside; larger, outside. Equality is the on-circle certificate.

Substitute the point into the left side and compare with the right: $(7 - 3)^{2} + (2 + 1)^{2} = 4^{2} + 3^{2} = 16 + 9 = 25.$ Right side is $25$ . Since $25 = 25$ , the point's distance from the centre is exactly $r$ . Answer: Yes, $(7, 2)$ lies on the circle.

Recall Solution 2.3

WHY find $r$ first: the point is on the rim, so its distance to the centre is the radius. Use the Distance Formula $d = (x_{2} - x_{1})^{2} + (y_{2} - y_{1})^{2}$ with the two points being centre $(2, - 4)$ and rim point $(6, - 1)$ : $r = (6 - 2)^{2} + (- 1 - (- 4))^{2} = 4^{2} + 3^{2} = 16 + 9 = 25 = 5.$ Now write the equation with $h = 2, k = - 4, r^{2} = 25$ : $(x - 2)^{2} + (y + 4)^{2} = 25$

Level 3 — Analysis

Now the standard form is hidden or partly given; you must uncover it.

Figure s02 — a lavender circle centred at the coral dot $(1, 2)$ dips down and just touches the horizontal $x$ -axis at the single slate point $(1, 0)$ . A mint vertical segment joins the centre straight down to that touch point; its length is the height $2$ of the centre above the axis, which equals the radius $r = 2$ .

Recall Solution 3.1

WHAT "tangent to the $x$ -axis" means: the circle just kisses the line $y = 0$ at one point. As the figure shows, the shortest distance from the centre down to that line is the radius.

The centre is at height $k = 2$ above the $x$ -axis. The straight-down distance to $y = 0$ is $∣2 - 0∣ = 2$ . So $r = 2$ , giving $r^{2} = 4$ . $(x - 1)^{2} + (y - 2)^{2} = 4$ WHY vertical distance: tangency to a horizontal line means the radius drawn to the touch-point is vertical, so its length is just the gap in the $y$ -coordinate.

Recall Solution 3.2

This is general form. To read a centre and radius we must return to standard form using Completing the Square (defined at the top: add half-the-linear-coefficient, squared).

Group $x$ and $y$ terms: $(x^{2} - 6 x) + (y^{2} + 8 y) = 11.$ Complete each square. For $x^{2} - 6 x$ , half of $- 6$ is $- 3$ ; its square is $9$ . For $y^{2} + 8 y$ , half of $8$ is $4$ ; its square is $16$ . Add these to both sides: $(x^{2} - 6 x + 9) + (y^{2} + 8 y + 16) = 11 + 9 + 16.$ WHY adding $9$ and $16$ creates perfect squares: a perfect square expands as $(x - a)^{2} = x^{2} - 2 a x + a^{2}$ . Compare with $x^{2} - 6 x$ : the middle term $- 6 x$ matches $- 2 a x$ only when $a = 3$ , and the missing piece to complete the pattern is the constant $a^{2} = 9$ . So $x^{2} - 6 x + 9$ is exactly $(x - 3)^{2}$ . Same story for $y$ : $y^{2} + 8 y$ needs $a = 4$ , whose square $16$ finishes it into $(y + 4)^{2}$ . We add these to the left to build the squares, so to keep the equation balanced we must add the identical amounts to the right. Fold into squares: $(x - 3)^{2} + (y + 4)^{2} = 36.$ Answer: centre $(3, - 4)$ , radius $36 = 6$ .

Recall Solution 3.3

WHY these two facts: the centre is the midpoint of a diameter, and the radius is half the diameter's length.

Centre = midpoint: $(\frac{- 1 + 5}{2}, \frac{2 + 10}{2}) = (2, 6) .$ Radius = half the distance $A B$ (using the Distance Formula): $A B = (5 - (- 1))^{2} + (10 - 2)^{2} = 6^{2} + 8^{2} = 36 + 64 = 100 = 10,$ so $r = \frac{10}{2} = 5$ and $r^{2} = 25$ . $(x - 2)^{2} + (y - 6)^{2} = 25$

Level 4 — Synthesis

Combine several ideas: geometry conditions, systems, distances.

Recall Solution 4.1

WHAT the trick is: every point on the circle is distance $r$ from the unknown centre $(h, k)$ . Setting pairs of these distances equal removes $r$ and $r^{2}$ cleanly.

WHY equal distances pin the centre: if the centre is equally far from two given points, it must lie on the perpendicular bisector of the segment joining them — the set of all points equidistant from both ends. Intersecting two such bisectors fixes the centre exactly. And there is a shortcut here: when two points share an $x$ -value, the segment between them is vertical, so its perpendicular bisector is the horizontal line through their midpoint's $y$ ; when two points share a $y$ -value, the bisector is the vertical line through their midpoint's $x$ . That is why the algebra below reduces to simply averaging coordinates.

Equal distance from $(1, 1)$ and $(1, 7)$ — both have $x = 1$ , so the centre lies on the horizontal line halfway between $y = 1$ and $y = 7$ : $k = \frac{1 + 7}{2} = 4.$ Equal distance from $(1, 7)$ and $(5, 7)$ — both have $y = 7$ , so the centre lies halfway between $x = 1$ and $x = 5$ : $h = \frac{1 + 5}{2} = 3.$ So the centre is $(3, 4)$ . Find $r$ using any of the three points, say $(1, 1)$ , via the Distance Formula: $r = (1 - 3)^{2} + (1 - 4)^{2} = 4 + 9 = 13 .$ $(x - 3)^{2} + (y - 4)^{2} = 13$ Check with $(5, 7)$ : $(5 - 3)^{2} + (7 - 4)^{2} = 4 + 9 = 13$ ✓.

Recall Solution 4.2

WHY horizontal distance now: tangency to a vertical line means the touching radius is horizontal, so its length is the gap in the $x$ -coordinates. $r = ∣8 - 3∣ = 5 \Rightarrow r^{2} = 25.$ $(x - 3)^{2} + (y + 2)^{2} = 25$

Level 5 — Mastery

The hardest blends: unknown radius from a condition, or two constraints at once.

Figure s03 — two lavender circles both hug the two positive axes, so each has its centre on the dashed grey line $y = x$ at a point $(r, r)$ . The small mint circle has radius $2$ (centre $(2, 2)$ ) and the large lavender circle has radius $10$ (centre $(10, 10)$ ); both pass through the coral point $(4, 2)$ , showing two valid solutions.

Recall Solution 5.1

WHAT tangency to both axes forces: if the circle touches the $x$ -axis, its distance to $y = 0$ is $r$ , so the centre's height is $r$ . If it touches the $y$ -axis, its distance to $x = 0$ is $r$ , so the centre's horizontal position is $r$ . In the first quadrant both are positive, so the centre is $(r, r)$ .

WHY the picture matters: as the figure shows, the centre sits equally far from both axes, so it lies on the line $y = x$ at $(r, r)$ .

Use the through-point $(4, 2)$ ; its distance to $(r, r)$ equals $r$ : $(4 - r)^{2} + (2 - r)^{2} = r^{2} .$ Expand each square: $(16 - 8 r + r^{2}) + (4 - 4 r + r^{2}) = r^{2} .$ Collect terms: $2 r^{2} - 12 r + 20 = r^{2} \Rightarrow r^{2} - 12 r + 20 = 0.$ Factor: $(r - 2) (r - 10) = 0$ , so $r = 2$ or $r = 10$ . Both are valid (both positive, both give first-quadrant circles through $(4, 2)$ ).

$(x - 2)^{2} + (y - 2)^{2} = 4 or (x - 10)^{2} + (y - 10)^{2} = 100$ Check small one: $(4 - 2)^{2} + (2 - 2)^{2} = 4 + 0 = 4$ ✓. Check large one: $(4 - 10)^{2} + (2 - 10)^{2} = 36 + 64 = 100$ ✓.

Recall Solution 5.2

Radius from the origin to $(6, 8)$ via the Distance Formula: $r = 6^{2} + 8^{2} = 36 + 64 = 100 = 10,$ so the circle is $x^{2} + y^{2} = 100$ . WHY negating both coordinates gives the opposite end: two ends of a diameter are the farthest-apart pair of rim points, and the centre is their exact midpoint. When the centre is the origin $(0, 0)$ , the midpoint condition says $\frac{6 + x ^{'}}{2} = 0$ and $\frac{8 + y ^{'}}{2} = 0$ for the opposite end $(x^{'}, y^{'})$ . Solving gives $x^{'} = - 6$ and $y^{'} = - 8$ — that is, the opposite point is the mirror image of $(6, 8)$ through the origin: same distance out, exactly the opposite direction. Negating both coordinates is precisely that reflection through the centre. $(6, 8) ⟶ (- 6, - 8) .$ Check it is on the circle: $(- 6)^{2} + (- 8)^{2} = 36 + 64 = 100$ ✓. Answer: $r = 10$ ; opposite point $(- 6, - 8)$ .

Recall Solution 5.3

Set up the centre. On the line $y = x$ the centre has equal coordinates, so write it as $(a, a)$ with $a > 0$ (first quadrant).

WHY tangency to the $y$ -axis fixes the radius. The $y$ -axis is the vertical line $x = 0$ . Tangency means the circle just touches this line at a single point, and the radius drawn to that touch-point runs straight across, perpendicular to the line — i.e. horizontally. Its length is therefore just the horizontal gap between the centre and the line: the horizontal gap is $∣ a - 0∣ = a$ . So the radius is $r = a$ .

Through-point condition at $(2, 0)$ : its distance to the centre $(a, a)$ must equal $r = a$ , so square both sides: $(2 - a)^{2} + (0 - a)^{2} = a^{2} .$ Expand using $(2 - a)^{2} = 4 - 4 a + a^{2}$ and $(0 - a)^{2} = a^{2}$ : $(4 - 4 a + a^{2}) + a^{2} = a^{2} \Rightarrow a^{2} - 4 a + 4 = 0 \Rightarrow (a - 2)^{2} = 0.$ This is a perfect square, giving the single root $a = 2$ (the one first-quadrant circle that fits). Then $r = a = 2$ . $(x - 2)^{2} + (y - 2)^{2} = 4$ Check: $(2 - 2)^{2} + (0 - 2)^{2} = 0 + 4 = 4$ ✓, and the distance from centre $(2, 2)$ to the $y$ -axis is $2 = r$ ✓.

Connections

Distance Formula — the radius is always a distance; used in almost every solution here.
Pythagorean Theorem — the engine inside the distance formula.
Completing the Square — needed in Exercise 3.2 to un-hide the standard form.
General Form of Circle — the expanded shape you convert from.
Equation of Tangent to Circle — next step after mastering tangency-to-a-line.
Parametric Equations of Circle — an alternative description of the same rim points.
Conic Sections — circles as the balanced special case of ellipses.

Recall Quick self-check ladder

L1 skill ::: read $(h, k)$ and $r$ with correct sign flip. L2 skill ::: build the equation; find $r$ as a distance when a rim point is given. L3 skill ::: uncover the centre via tangency-to-axis, completing the square, or a diameter's midpoint. L4 skill ::: pin the centre by equating distances / using perpendicular-bisector logic. L5 skill ::: turn a geometric condition into a quadratic in $r$ and keep all valid roots.