NIMCET 2019 Question Paper with solutions

Question : 1 of 120

Marks: +1, -0

For two circles

x^2 + y^2

= 16 and

x^2 + y^2

— 2y = 0, there is / are

Solution:

Given,

x^{2}+y^{2}=16

Center

(0,0),r_i=4

x^{2}+y^{2}-2y=0

(x-0)^{2}+(y-1)^{2}=1

Center

(0,1),r_2=1

Distance between their centre

=√{(0-0)^{2}+(1-0)^{2}}

=√{0+1}=1

|r_{1}-r_{2}|=4-1=3

c_{1}c_{2}< |r_{1}-r_{2}|

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