Question

Find the equation of the circle passing through the points $(0,-1)$ and $(2,0)$ and whose centre lies on the line $3x+y=5$.

Answer 1

Let A= (0,-1) B= (2,0)

The midpoint of segment AB is M = $\frac{A+B}{2}$

$=(\frac{0+2}{2},\frac{0-1}{2})$

$M=(1,-\frac{1}{2})$

equation of $\perp$ bisector of AB is

$y+\frac{1}{2}=2(x-1)$

$2x-y=\frac{5}{2}$

$4x-2y=5$ ________ (1)

The center must be intersaction of equation (1) and 3x+y=5 _______ (2) in which center lie.

4x-2y=5 radius = $\sqrt{(\frac{3}{2}-0{)}^2+(\frac{1}{2}+1{)}^2}=\frac{3}{\sqrt{2}}$

6x+2y=10

_________

10x= 15

$[x=\frac{3}{2}]$ and $[y=\frac{1}{2}]$

equation of circle with center $(\frac{3}{2},\frac{1}{2})$ & radius

$(x-\frac{3}{2}{)}^2+(y-\frac{1}{2}{)}^2=\frac{9}{4}$ _____ (3)