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10-1.Circle and System of Circles
hard
The point of contact of the given circles ${x^2} + {y^2} - 6x - 6y + 10 = 0$ and ${x^2} + {y^2} = 2$, is
A
$(0, 0)$
B
$(1, 1)$
C
$(1, -1)$
D
$(-1, -1)$
Solution
(b) ${x^2} + {y^2} – 6x – 6y + 10 = 0$….$(i)$
${x^2} + {y^2} = 2$….$(ii)$
$ \Rightarrow – 6x – 6y + 12 = 0$
or $x + y – 2 = 0$….$(iii)$
$ \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.
Standard 11
Mathematics
