UPSC NDA Math Model Paper 4

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: $(x - h)^2 + (y - k)^2 = r^2$ Here, we have h = 0, k = c and r = 5 $\Rightarrow (x-0)^{2}+(y-c)^{2}=5^{2}$ .......(1) ∵ The circle also passes through the point (3, 2) So, x = 3 and y = 2 will satisfy the equation (1) $\Rightarrow (3-0)^{2}+(2-c)^{2}=25$ $\Rightarrow (2-c)^{2}=16$ $\Rightarrow (2-c)=\pm 4$ 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 $\Rightarrow x^{2}+(y+2)^{2}=25$ $\Rightarrow x^{2}+y^{2}+4 y-21=0$ So, the equation of the required circle is $x^{2}+y^{2}+4 y-21=0$ Hence, option D is the correct answer.