Circles Through Given Points & Concyclicity

Given the endpoints of a diameter $(x_1,y_1)$ and $(x_2,y_2)$, the circle is

• This is the constructive use of the diameter form — every point sees the diameter at $90^\circ$, so $\vec{PA}\cdot\vec{PB}=0$.
• Equivalently, find the centre as the midpoint $\left(\tfrac{x_1+x_2}{2},\tfrac{y_1+y_2}{2}\right)$ and the radius as half the distance $\tfrac12\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}$, then use standard form.

To find the circle through three points, start from the general form with unknowns:

• Substitute each of the three points to get three linear equations in $D,E,F$; solve the system.
• The centre is then $\left(-\tfrac D2,-\tfrac E2\right)$ and the radius is $\sqrt{\tfrac{D^2}{4}+\tfrac{E^2}{4}-F}$.
• Subtracting pairs of the equations eliminates the $x^2+y^2$ terms and gives the two perpendicular-bisector lines; their intersection is the centre — a useful shortcut.

After solving the three-point system $x^2+y^2+Dx+Ey+F=0$:

• Centre $=\left(-\tfrac D2,-\tfrac E2\right)$.
• Radius is most safely found as the distance from the centre to any one of the three given points: $r^2=(x_0-h)^2+(y_0-k)^2$. This avoids sign errors in $\sqrt{g^2+f^2-c}$.
• When the question only asks a comparison ("is $r>60$?"), compare $r^2$ against the threshold squared — no square root needed.

Given two points the circle passes through and a line the centre lies on:

• The centre is equidistant from the two points $A,B$, so it lies on the perpendicular bisector of $AB$. Form that bisector (equate $CA^2=CB^2$).
• Intersect the perpendicular bisector with the given line to get the centre $(h,k)$.
• The radius is the distance from $(h,k)$ to either point; write the circle in standard form, then expand to general form to match the options.

Concyclicity test: four points are concyclic $\iff$ the fourth lies on the circle through the first three.

• Find the circle through three of the points (general-equation system), then substitute the fourth point; it must give $0$.
• If the fourth point has an unknown coordinate $(0,k)$, substitution produces a quadratic in $k$ — its roots are the value(s) that make all four concyclic.
• Once the circle is known, its diameter is $2r=2\sqrt{g^2+f^2-c}$.

If $S\equiv x^2+y^2+\ldots=0$ is a circle and $L\equiv ax+by+c=0$ a line cutting it in a chord, then

• Apply the extra condition to fix $\lambda$: e.g. "the chord is a diameter" means the new circle's centre lies on $L$ — set the centre $\left(-\tfrac{\lambda a}{2},-\tfrac{\lambda b}{2}\right)$ on $L$ and solve for $\lambda$.
• Similarly $S_1+\lambda S_2=0$ (two circles) gives the family through their common points; $\lambda=-1$ gives the radical axis (common chord).

For a triangle with a right angle, the circumcentre (centre of the circle through all three vertices) is the midpoint of the hypotenuse, and the circumradius is half the hypotenuse.

• Spotting the right angle: two of the bounding lines are perpendicular (e.g. $x=\text{const}$ and $y=\text{const}$, or slopes whose product is $-1$).
• Then set the midpoint of the hypotenuse equal to the given circumcentre and solve for the unknown parameter.