Given: Circle passes through (3, 4) and touch y-axis at origin, this means it also passes though origin O(0, 0).
To Find: Equation of Circle
General equation of Circle: x2 + y2 + 2gx + 2fy + c = 0
As circle passes through (0, 0). This point will satisfy the equation.
Therefore,
(0)2 + (0)2 + 2g(0) + 2f(0) + c = 0
c = 0
Now, we have the equation of circle as,
x2 + y2 + 2gx + 2fy = 0
Now, since the centre is on the x-axis, y coordinate of centre = 0. i.e. f = 0.
Therefore,
x2 + y2 + 2gx = 0
It is also given that this circle passes through (3, 4). So,
(3)2 + (4)2 + 2g(3) = 0
9 + 16 + 6g = 0
25 + 6g = 0
6g = -25
g = \(-\cfrac{25}6\)
Hence, we have the equation as,
