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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).

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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

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