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

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 $$1$$, $${ g }_{ 1 }{ g }_{ 2 }=-{ a }^{ 2 }$$) $$\Rightarrow { c }^{ 2 }=2{ a }^{ 2 }+{ a }^{ 2 }{ m }^{ 2 }$$$$\Rightarrow { c }^{ 2 }={ a }^{ 2 }\left( 2+{ m }^{ 2 } \right)$$Maths

Suggest Corrections

0

Similar questions
View More

People also searched for
View More