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10-1.Circle and System of Circles
hard
A circle passes through the origin and has its centre on $y = x$. If it cuts ${x^2} + {y^2} - 4x - 6y + 10 = 0$ orthogonally, then the equation of the circle is
A
${x^2} + {y^2} - x - y = 0$
B
${x^2} + {y^2} - 6x - 4y = 0$
C
${x^2} + {y^2} - 2x - 2y = 0$
D
${x^2} + {y^2} + 2x + 2y = 0$
Solution
(c) Let the required circle be
${x^2} + {y^2} + 2gx + 2fy + c = 0$….$(i)$
This pases through $(0, 0)$, therfore $c = 0$.
The centre $( – g,\; – f)$ of $(i)$ lies on $y = x$, hence $g = f$.
Since $(i)$ cuts the circle ${x^2} + {y^2} – 4x – 6y + 10 = 0$ orthogonally, therefore
$2( – 2g – 3f) = c + 10 \Rightarrow – 10g = 10$
$ \Rightarrow g = f = – 1$
Hence the required circle is ${x^2} + {y^2} – 2x – 2y = 0$.
Standard 11
Mathematics
