  Question

CirclesNumber of common tangentsI:$$x^{2}+y^{2}=4$$$$x^{2}+y^{2}-8x+12=0$$a) $$0$$II:$$x^{2}+y^{2}=1$$$$x^{2}+y^{2}-2x-6y+6=0$$b) $$1$$III:$$x^{2}+y^{2}=16$$$$x^{2}+y^{2}-8x+6y-56=0$$c) $$4$$IV: $$x^{2}+y^{2}-2x-6y+9=0$$$$x^{2}+y^{2}+6x-2y+1=0$$d)$$3$$

A
a,b,c,d  B
d,c,b,c  C
c,b,a,d  D
d,c,b,a  Solution

The correct option is A $$d,c,b,c$$$$I.\ S_{1}:x^{2}+y^{2}=4$$  and  $$S_{2}:x^{2}+y^{2}-8x+12=0$$      $$C_{1}\equiv (0,0)$$                     $$C_{2}\equiv (4,0)$$       $$r_{1}=2$$                             $$r_{2}=2$$$$S_{1}$$ and $$S_{2}$$ Touch each other externally, So no of common tangent are $$3.$$$$II.\ S_{1}:x^{2}+y^{2}=1$$  and  $$S_{2}:x^{2}+y^{2}-2x-6y+6=0$$     $$C_{1}\equiv (0,0)$$                 $$C_{2}\equiv (1,3)$$     $$r_{1}=1$$                         $$r_{2}=2$$$$C_{1}C_{2}=\sqrt{10}$$  and  $$r_{1}+r_{2}=3$$$$C_{1}C_{2}>r_{1}+r_{2}$$$$S_{1}$$ & $$S_{2}$$ does not intersect each other. So no of common tangents are $$4$$.$$III.\ S_{1}:x^{2}+y^{2}=16$$  and  $$S_{2}:x^{2}+y^{2}-8x+6y+56=0$$     $$C_{1}\equiv (0,0)$$                 $$C_{2}\equiv (4,3)$$     $$r_{1}=4$$                         $$r_{2}=9$$$$C_{1}C_{2}=5$$   and   $$r_{1}+r_{2}=13$$$$r_{2}-r_{1}=5$$$$C_{1}C_{2}=r_{2}-r_{1}$$$$S_{1}$$ and $$S_{2}$$ Touches internally so no of common tangent are $$1$$.$$IV.\ x^{2}+y^{2}-2x-6x+9=0$$       $$C_{1}\equiv (1,3)$$                     $$r_{1}=1$$$$x^{2}+y^{2}+6x-2y+1=0$$$$C_{2}\equiv (-3,1)$$$$r_{2}=3$$$$C_{1}C_{2}>r_{1}r_{2}$$$$S_{1}$$ and $$S_{2}$$ are not intersection so, no of common tangent is $$4.$$So, $$d,c,b,c$$Maths

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