Question

The equation of a circle of radius $$2$$ touching the circles $${ x }^{ 2 }+{ y }^{ 2 }-4\left| x \right| =0$$ is

A
x2+y2+23y+2=0
B
x2+y2+43y+8=0
C
x2+y243y+8=0
D
None of these

Solution

The correct options are B $${ x }^{ 2 }+{ y }^{ 2 }+4\sqrt { 3 } y+8=0$$ C $${ x }^{ 2 }+{ y }^{ 2 }-4\sqrt { 3 } y+8=0$$The given circle are $${ x }^{ 2 }+{ y }^{ 2 }-4x=0,x>0$$i.e., $${ \left( x-2 \right) }^{ 2 }+{ y }^{ 2 }={ 2 }^{ 2 },x>0.$$and, $${ x }^{ 2 }+{ y }^{ 2 }+4x=0,x<0$$i.e., $${ \left( x+2 \right) }^{ 2 }+{ y }^{ 2 }={ 2 }^{ 2 },x<0.$$So, the centres of the given circles are $$(-2,0)$$ and $$(2,0)$$ and both having same radius $$2$$ units.Clearly, from the figure, the centers of the circles forms an equilateral triangle of side $$4$$ units.Height of this equilateral triangle $$=\cfrac{\sqrt{3}}{2} \times 4 = 2\sqrt{3}=\sqrt{12}$$So, the centres of the required circle are at $$\left( 0,\sqrt { 12 } \right)$$ and $$\left( 0,-\sqrt { 12 } \right).$$$$\therefore$$ Equations of the required circle are $${ \left( x-0 \right) }^{ 2 }+{ \left( y\mp \sqrt { 12 } \right) }^{ 2 }={ 2 }^{ 2 }$$i.e., $${ x }^{ 2 }+{ y }^{ 2 }+4\sqrt { 3 } y+8=0.$$ and $${ x }^{ 2 }+{ y }^{ 2 }-4\sqrt { 3 } y+8=0.$$Mathematics

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