Here, we have to find equation of the circle of radius 5 units whose centre lies on the y -axis below the origin and which also passes through the point (3, 2) ∵ Let the centre of the circle be (0, c) As we know that, the equation of circle with centre at (h, k) and radius r units is given by: (x−h)2+(y−k)2=r2 Here, we have h = 0, k = c and r = 5 ⇒(x−0)2+(y−c)2=52 .......(1) ∵ The circle also passes through the point (3, 2) So, x = 3 and y = 2 will satisfy the equation (1) ⇒(3−0)2+(2−c)2=25 ⇒(2−c)2=16 ⇒(2−c)=±4 If (2 - c) = 4 then c = - 2 and if (2 - c) = - 4 then c = 6 But ∵ the centre lies on the y -axis below the origin, so c = - 2 So, the centre of the circle is at (0, - 2) By substituting c = - 2 in equation (1), we get ⇒x2+(y+2)2=25 ⇒x2+y2+4y−21=0 So, the equation of the required circle is x2+y2+4y−21=0 Hence, option D is the correct answer.