Circle Equation — Centre, Radius & Properties

A circle is the set of all points at a fixed distance $r$ (the radius) from a fixed point $C=(h,k)$ (the centre).

• Standard (centre–radius) form: a point $(x,y)$ is on the circle when its distance to the centre equals $r$. Squaring the distance formula,

• A chord is a segment joining two points on the circle; the longest chord, passing through the centre, is a diameter $=2r$.
• The circle centred at the origin with radius $r$ is simply $x^2 + y^2 = r^2$.

Expanding $(x-h)^2+(y-k)^2=r^2$ gives the general form

• Centre $= (-g,\,-f)$ — minus half the coefficient of $x$ and of $y$.
• Radius $= \sqrt{g^2 + f^2 - c}$ (a real circle needs $g^2+f^2-c > 0$).
• Watch the leading coefficient: if the equation reads $Ax^2+Ay^2+\ldots=0$ with $A\neq 1$, divide through by $A$ first so the $x^2$ and $y^2$ coefficients are $1$ — otherwise the centre/radius formulas give wrong numbers.

The circle with a diameter from $(x_1,y_1)$ to $(x_2,y_2)$ is

• It comes from the angle-in-a-semicircle fact: a point $P=(x,y)$ is on the circle exactly when $\vec{PA}\perp\vec{PB}$, i.e. their dot product is zero.
• Recognising a circle ALREADY in this factored shape $\big((x-p)(x-q)+(y-s)(y-t)=0\big)$ hands you the diameter endpoints $(p,s)$ and $(q,t)$ — and the centre is their midpoint $\big(\tfrac{p+q}{2},\tfrac{s+t}{2}\big)$.

For a circle, the chord it cuts on an axis is found by zeroing the other coordinate:

• y-axis intercept: put $x=0$; the equation becomes a quadratic in $y$. Its two roots $y_1,y_2$ are where the circle meets the y-axis, and the intercept length is $|y_1 - y_2|$.
• x-axis intercept: put $y=0$ and read the gap between the roots in $x$ the same way.
• In general-form symbols, the x-axis intercept length is $2\sqrt{g^2-c}$ and the y-axis intercept length is $2\sqrt{f^2-c}$ (real only when the bracket is positive).

A fundamental circle property: the perpendicular from the centre to a chord bisects the chord (and, conversely, the line from the centre to a chord's midpoint is perpendicular to the chord).

• Midpoint of a chord on a line $L$: drop a perpendicular from the centre $C$ to $L$; the foot of that perpendicular is the midpoint. Build the line through $C$ with slope $=-1/(\text{slope of }L)$ and intersect it with $L$.
• Length of a chord at perpendicular distance $d$ from the centre: $2\sqrt{r^2 - d^2}$.
• The same construction locates a point of TANGENCY: the foot of the perpendicular from the centre to a tangent line IS the contact point (distance from the centre $=r$). General recipe for the foot of the perpendicular from $(h,k)$ to $ax+by+c=0$: $\big(h-a\,t,\ k-b\,t\big)$ with $t=\dfrac{ah+bk+c}{a^2+b^2}$.

Tangency to a line = distance from centre equals radius.

• Touches the x-axis $\iff$ $|k| = r$ (the centre's height equals the radius). Touches the y-axis $\iff |h| = r$.
• Touches BOTH axes in the first quadrant $\iff$ centre $=(r,r)$, so the equation is $(x-r)^2+(y-r)^2=r^2$.
• Touches a general line $ax+by+c=0$ $\iff$ $\dfrac{|ah+bk+c|}{\sqrt{a^2+b^2}} = r$.
• Touches TWO intersecting lines $\iff$ the centre is equidistant from both, i.e. it lies on one of their angle bisectors $\dfrac{a_1x+b_1y+c_1}{\sqrt{a_1^2+b_1^2}}=\pm\dfrac{a_2x+b_2y+c_2}{\sqrt{a_2^2+b_2^2}}$; the coordinate axes are the case where the bisectors are $y=\pm x$, which is why the centre is $(r,\pm r)$. The bisector pins the centre to a one-parameter family; one more condition (a point on the circle, a stated distance of the contact point from the origin) fixes it. The contact point is the foot of the perpendicular from the centre to the line.

For circles with centres $C_1,C_2$, radii $r_1,r_2$, and centre distance $d=|C_1C_2|$:

• Two distinct intersection points $\iff |r_1 - r_2| < d < r_1 + r_2$.
• Touch externally (one common point) $\iff d = r_1+r_2$; touch internally $\iff d = |r_1-r_2|$.
• Lie outside each other (no common point) $\iff d > r_1+r_2$; one inside the other $\iff d < |r_1-r_2|$.

A circle through the origin making intercepts $a$ on the x-axis and $b$ on the y-axis passes through $(0,0),(a,0),(0,b)$:

• Its general form is $x^2+y^2-ax-by=0$ (the constant term is $0$ because it passes through the origin).
• Centre $=\left(\tfrac a2,\tfrac b2\right)$ — the midpoint of the axis-crossings, since $(a,0)$ and $(0,b)$ are ends of a diameter (the angle at the origin is a right angle).
• To test which line the centre lies on, substitute $\left(\tfrac a2,\tfrac b2\right)$ into each candidate.