CameraIcon
CameraIcon
SearchIcon
MyQuestionIcon


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
loader
B
x2+y2+43y+8=0
loader
C
x2+y243y+8=0
loader
D
None of these
loader

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 $${ x }^{ 2 }+{ y }^{ 2 }-4\sqrt { 3 } y+8=0.$$

387302_290087_ans_6928dc1949214e09b63baac07a59ee0d.png

Mathematics

Suggest Corrections
thumbs-up
 
0


similar_icon
Similar questions
View More


similar_icon
People also searched for
View More



footer-image