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10-1.Circle and System of Circles
hard
The equation of a circle that intersects the circle ${x^2} + {y^2} + 14x + 6y + 2 = 0$orthogonally and whose centre is $(0, 2)$ is
A
${x^2} + {y^2} - 4y - 6 = 0$
B
${x^2} + {y^2} + 4y - 14 = 0$
C
${x^2} + {y^2} + 4y + 14 = 0$
D
${x^2} + {y^2} - 4y - 14 = 0$
Solution
(d) In circle, ${x^2} + {y^2} + 14x + 6y + 2 = 0$
$g = 7,\;f = 3,\;c = 2$
Centre of circle $( – g,\; – f) = (0,\;2)$, (Given)
For orthogonally intersection, $2gg' + 2ff' = c + c'$
$0 – 12 = 2 + c' \Rightarrow c' = – 14$
Put the values, in equation ${x^2} + {y^2} + 2g'x + 2f'x + c' = 0$.
$ \Rightarrow {x^2} + {y^2} + 0 – 4y – 14 = 0 $
$\Rightarrow {x^2} + {y^2} – 4y – 14 = 0$.
Standard 11
Mathematics
