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Question

The condition that the circles which passes through the points $$(0,\mathrm{a})$$ , $$(0,- \mathrm{a})$$ and touch the line $$\mathrm{y}=\mathrm{m}\mathrm{x}+\mathrm{c}$$ will cut orthogonally is


A
c2=a2(1+m2)
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B
c2=a2(2+m2)
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C
c2=a2(3+m2)
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D
c2=a2(4+m2)
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Solution

The correct option is B $$\mathrm{c}^{2}=\mathrm{a}^{2}(2+\mathrm{m}^{2})$$
Let the centers be $${ C }_{ 1 }\left( -{ g }_{ 1 },0 \right) $$ & $${ C }_{ 2 }\left( -{ g }_{ 2 },0 \right) $$
$$\because $$They cut each other orthogonally and pass through $$\left( 0,a \right) $$ and $$\left( 0,-a \right) $$ their equations are
$${ S }_{ 1 }={ x }^{ 2 }+{ y }^{ 2 }+2{ g }_{ 1 }x-{ a }^{ 2 }=0,{ S }_{ 2 }={ x }^{ 2 }+{ y }^{ 2 }+2{ g }_{ 2 }x-{ a }^{ 2 }=0$$
Then $$2{ g }_{ 1 }{ g }_{ 2 }+2{ f }_{ 1 }{ f }_{ 2 }={ c }_{ 1 }+{ c }_{ 2 }...$$ gives
$$2{ g }_{ 1 }{ g }_{ 2 }+2\left( 0 \right) \left( 0 \right) =-{ a }^{ 2 }-{ a }^{ 2 }....2{ g }_{ 1 }{ g }_{ 2 }=-2.{ a }^{ 2 }$$
So that $${ g }_{ 1 }{ g }_{ 2 }=-{ a }^{ 2 }...1$$
Now if a circle $${ x }^{ 2 }+{ y }^{ 2 }+2gx-{ a }^{ 2 }=0$$ with center $$c\left( -{ g }_{ 1 },0 \right) $$ and $$r=\sqrt { { g }^{ 2 }+{ a }^{ 2 } } $$ where g is a parameter touches the line $$y=mx+c$$ i.e., $$mx-1y+c=0$$ then $$CP=r$$, gives
$$\cfrac { \left[ m\left( -g \right) -\left( 0 \right) +c \right]  }{ \sqrt { { m }^{ 2 }+1 }  } =\sqrt { { g }^{ 2 }+{ a }^{ 2 } } $$
$$\Rightarrow c-mg=\sqrt { { g }^{ 2 }+{ a }^{ 2 } } \sqrt { { m }^{ 2 }+1 } $$
Squaring on both sides
$${ c }^{ 2 }-2cmg+{ m }^{ 2 }{ g }^{ 2 }=\left( { g }^{ 2 }+{ a }^{ 2 } \right) { m }^{ 2 }+\left( { g }^{ 2 }+{ a }^{ 2 } \right) $$
$${ g }^{ 2 }+2cmg+\left[ { a }^{ 2 }{ m }^{ 2 }+{ a }^{ 2 }-{ c }^{ 2 } \right] =0....2$$
This is a quadratic equation in g
$${ g }_{ 1 }{ g }_{ 2 }=\left[ \cfrac { \left( { a }^{ 2 }{ m }^{ 2 }+{ a }^{ 2 }-{ c }^{ 2 } \right)  }{ 1 }  \right] ={ a }^{ 2 }$$ ($$\because $$ from $$1$$, $${ g }_{ 1 }{ g }_{ 2 }=-{ a }^{ 2 }$$)
$$\Rightarrow { c }^{ 2 }=2{ a }^{ 2 }+{ a }^{ 2 }{ m }^{ 2 }$$
$$\Rightarrow { c }^{ 2 }={ a }^{ 2 }\left( 2+{ m }^{ 2 } \right) $$

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