Find the equation of the circle of radius 5 cm, whose centre lies on the y - axis and which passes through the point (3, 2).
The general form of the equation of a circle is:
(x - h)2 + (y - k)2 = r2
Where, (h, k) is the centre of the circle.
r is the radius of the circle.
Since, centre lies on Y - axis, ∴ it’s X - coordinate = 0, i.e.h = 0
Hence, (0, k) is the centre of the circle.
Substituting the given values in general form of the equation of a circle we get,
⇒ (3 - 0)2 + (2 - k)2 = 52
⇒ (3)2 + (2 - k)2 = 25
⇒ 9 + (2 - k)2 = 25
⇒ (2 - k)2 = 25 - 9 = 16
Taking square root on both sides we get,
⇒ 2 - k = ±4
⇒ 2 - k = 4 & 2 - k = - 4
⇒ k = 2 - 4 & k = 2 + 4
⇒ k = - 2 & k = 6
∴ Equation of circle when k = - 2 is:
x2 + (y + 2)2 = 25
Equation of circle when k = 6 is:
x2 + (y - 6)2 = 25
Ans: Equation of circle when k = - 2 is:
x2 + (y + 2)2 = 25
Equation of circle when k = 6 is:
x2 + (y - 6)2 = 25
