Given centre of circle, C(1, 0) and external point P(3, 3/2).
(x - h)2 + (y - k)2 = r2
(1 - 3)2 + (0 - 3 / 2) = r2
(-2)2 + (-3 / 2)2 = r2
r2 = 25 / 4
r = (5 / 2) units
Therefore the equation of the circle is,
(x - 1)2 + (y - 0)2 = (5 / 2)2
x2 -2x + 1 + y2 = (25 / 4)
On simplifying we get,
4x2 + 4y2 - 8x - 21 = 0
The question should be "Find also the equation of the equal circle..." otherwise, there is insufficient information to be able to determine the equation of the circle.
Note:
When two circles touch externally,
Distance between their centres = sum of their radii
Since the circles are equal, their radii are the same(= 5 / 2 units).
Since they touch externally at P and their radii are the same, P is the midpoint of the line joining the two circles.
Let the centre of the second circle be C(h, k)
Applying midpoint formula,
Px = (h + 1) / 2
3 = (h + 1) / 2
h + 1 = 6
h = 5
Py = (k + 0) / 2
(3 / 2) = (k + 0) / 2
k = 3
Therefore, the centre of the second circle C(h, k) = (5, 3)
The equation of the second circle is,
(x - h)2 + (y - k)2 = r2
(x - 5)2 + (y - 3)2 = (5 / 2)2
x2 + 25 - 10x + y2 + 9 - 6y = (25 / 4)
On simplifying we get,
4x2 + 4y2 - 40x - 24y + 111 = 0
