Question

Find the equation of the circle are which passes through the point ( 1 , -2) and (3, -4) and touched the axis of x

Answer 1

Since the circle touches x-axis therefore its equation is (x - h)${}^2$ + (y - k)${}^2$ = $k^2$ .(1)
(Rule (c), P. 850)
It passes through the given points.
$\therefore$ (1 - h)${}^2$ + (-2 - k)${}^2$ = $k^2$ ..(2)
(3 - h)${}^2$ + (-4 k)${}^2$ = $k^2$
Subtracting (2) and (3), we get h = k + 5
Putting in (2), we get $k^2$ + 12 k + 20 = 0
$\therefore$ k = -10, -2 and hence h = -5;3
$\therefore$ Centres are (-5, -10) or (3, -2)
$\therefore$ Circles are (x + 5)${}^2$ + (y + 10)${}^2$ = ${10}^2$
and (x - 3)${}^2$ + (y + 2)${}^2$ = $2^2$