Perpendicular bisector and circle through three points

Text

Consider the points $A(2;-1)$ and $B(-3;0)$:

• (a) find the perpendicular bisector of $AB$ using the locus definition;
• (b) find it as the perpendicular through the midpoint and compare;
• (c) write the equation of the circle through $A$, $B$ and the origin, and its radius.

Solution

(a) $d(P,A)=d(P,B)$: $(x-2)^2+(y+1)^2=(x+3)^2+y^2$. Expanding: $-4x+2y+5=6x+9\Rightarrow y=5x+2$.

(b) Midpoint $M\left(-\tfrac12;-\tfrac12\right)$; slope $AB=-\tfrac15$, perpendicular $5$: $y+\tfrac12=5\left(x+\tfrac12\right)\Rightarrow y=5x+2$. Same result. ✓

(c) Circle $x^2+y^2+Dx+Ey+F=0$ through the origin gives $F=0$. Through $B(-3;0)$: $D=3$. Through $A(2;-1)$: $E=11$. Hence $x^2+y^2+3x+11y=0$, centre $C\left(-\tfrac32;-\tfrac{11}{2}\right)$ and $R=\sqrt{\tfrac94+\tfrac{121}{4}}=\frac{\sqrt{130}}{2}\approx 5{,}70.$

$\boxed{x^2+y^2+3x+11y=0,\qquad C\left(-\tfrac32;-\tfrac{11}{2}\right),\qquad R=\tfrac{\sqrt{130}}{2}}$