question_answer

Let be a circle passing through \[\left( 2,0 \right)\] and \[\left( 0,2 \right)\] and having the small possible area. \[{{S}_{2}}\] is the circle touching \[{{S}_{1}}\] externally and passing through origin. \[{{S}_{3}}\] is a circle that touches and \[{{S}_{2}}\] both externally. If the area of each of \[{{S}_{1}}\] \[{{S}_{2}}\] and \[{{S}_{3}}\] is same, then

A) Equation of \[{{S}_{2}}\] is \[{{x}^{2}}+{{y}^{2}}+2x-2y=0\]

B) Equation of the smallest circle containing \[{{S}_{1}}\operatorname{and}\,{{S}_{2}}\,\operatorname{is}{{x}^{2}}+{{y}^{2}}=8\]

C) Coordinated of the Centre of \[{{S}_{3}}\] and \[\left( 2,-2 \right)\] or \[\left( -2,2 \right)\]

D) The circle intersecting \[S{{ }_{1}},{{S}_{2}}\,\,and\,\,{{S}_{3}}\] orthogonally has radius \[\sqrt{\frac{3}{2}}\]

Correct Answer: A

Solution :

(a)

\[{{S}_{1}}\]is the circle with gives points as diameter, so , \[{{S}_{1}}\]is \[{{x}^{2}}+{{y}^{2}}-2x-2y=0,\]which passes through origin.

Clearly, \[{{S}_{2}}\operatorname{is}\,{{x}^{2}}+{{y}^{2}}+2x-2y=0.\]

The smallest circle containing \[{{S}_{1}}\]and \[{{S}_{2}}\]has Centre at \[\left( 0,0 \right)\]and radius equal to diameter of\[{{S}_{1}}\]\[\left( \operatorname{or}\,{{S}_{2}} \right)\].

Now, Centre of \[{{S}_{3}}\] lies on \[x+y=0,\] let it be\[\sqrt{\left( h,-h \right)}\],

then \[\sqrt{{{\left( h-1 \right)}^{2}}+{{\left( -h-1 \right)}^{2}}}=2\sqrt{2}\Rightarrow h=\pm \sqrt{3}\]

\[\therefore \]Centre of \[{{S}_{3}}\]is \[\left( \sqrt{3},-\sqrt{3} \right)\operatorname{or}\left( -\sqrt{3},\sqrt{3} \right).\]

 

One of the equations of \[{{S}_{3}}\]is \[{{x}^{2}}+{{y}^{2}}-2\sqrt{3}x+2\sqrt{3}y+4=0.\]

Common tangents of \[{{S}_{1}}\]and \[{{S}_{2}}\]is \[x+y=0\]and that of \[{{S}_{1}}\]and \[{{S}_{3}}\]is

\[\left( \sqrt{3}-1 \right)x-\left( \sqrt{3}+1 \right)y-2=0.\]

So, the radical Centre intersecting \[{{S}_{1}}\],\[{{S}_{2}}\]and \[{{S}_{3}}\]orthogonally=\[\operatorname{OP}=\sqrt{\frac{2}{3}.}\]