Question

Find the equations to the circles which pass through the point :
(1, 2), (3, -4), and (5, -6).

Answer 1

Let $(h,k)$ be the centre of the circle and $r$ be the radius

The distance between each point and the centre will be equal to each other and also equal to the radius

$(1-h{)}^2+(2-k{)}^2=(3-h{)}^2+(-4-k{)}^2=(5-h{)}^2+(-6-k{)}^2=r^2$

For equality one

$5-2h-4k=25-6h+8k$

$h-3k=5$ ....... (1)

For second equality

$5-2h-4k=25-10h+36+12k$

$h-2k=7$ ......(2)

Solving equation (1) and (2)

$h=11,k=2$

and radius $r=\sqrt{(1-11{)}^2+(2-2{)}^2}$

$r=10$

So, the equation of circle is

$(x-11{)}^2+(y-2{)}^2=100$

$x^2+y^2-22x-4y+25=0$