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KEAM 2014 Math Paper

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Question : 115 of 120

Marks: +1, -0

A circle of radius

\sqrt{8}

is passing through origin and the point (4, 0). If the centre lies on the line

y = x,

then the equation of the circle is

$(x-2)^2+(y-2)^2=8$
$(x+2)^2+(y+2)^2=8$
$(x-3)^2+(y-3)^2=8$
$(x+3)^2+(y+3)^2=8$
$(x-4)^2+(y-4)^2=8$

Solution:

Let the equation of circle be

x^{2}+y^{2}+2 g x+2 f y+c=0,

whose centre

(-g,-f)

since, the circle passing through (0,0) and (4,0)

\therefore 0^{2}+0^{2}+2 g(0)+2 f(0)+c=0 \Rightarrow c=0

and

4^{2}+0^{2}+2 g(4)+2 f(0)+0=0

\Rightarrow \;\; 16+8 g=0 \Rightarrow g=-2

Also, the centre

(-g,-f)

lies on line

y=x

\therefore \;\; -f=-g \Rightarrow f=g

\therefore \;\; f=-2

Now, the equation of circle having centre (2,2) and radius

\sqrt{8}

is

(x-2)^{2}+(y-2)^{2}=(\sqrt{8})^{2}

\Rightarrow \;\; (x-2)^{2}+(y-2)^{2}=8

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