The point of contact of the given circles {x^ | vedclass

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Question

(b) ${x^2} + {y^2} - 6x - 6y + 10 = 0$&hellip;.$(i)$

${x^2} + {y^2} = 2$&hellip;.$(ii)$

$ \Rightarrow - 6x - 6y + 12 = 0$

or $x + y - 2 = 0$&

Vedclass · Accepted Answer

$(1, 1)$

Vedclass · Answer

(b) ${x^2} + {y^2} - 6x - 6y + 10 = 0$&hellip;.$(i)$

${x^2} + {y^2} = 2$&hellip;.$(ii)$

$ \Rightarrow - 6x - 6y + 12 = 0$

or $x + y - 2 = 0$&hellip;.$(iii)$

$ \Rightarrow {x^2} + {y^2} + 2xy = 4$ {from $(iii)$}

or $2xy = 2$ {from $(ii)$}

and $x - y = \sqrt {{{(x + y)}^2} - 4xy}$

$= \sqrt {4 - 4} = 0$

or $x = y$ and $x + y = 2$

$ \Rightarrow x = 1,\;y = 1$.

Trick : Required point must satisfy both the circles.

Obviously $(1, 1)$ satisfies both the equations simultaneously.

The point of contact of the given circles ${x^2} + {y^2} - 6x - 6y + 10 = 0$ and ${x^2} + {y^2} = 2$, is

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