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10-1.Circle and System of Circles
medium
If the two circles $2{x^2} + 2{y^2} - 3x + 6y + k = 0$ and ${x^2} + {y^2} - 4x + 10y + 16 = 0$ cut orthogonally, then the value of $k$ is
A
$41$
B
$14$
C
$4$
D
$0$
Solution
(c) Given circles are
$2{x^2} + 2{y^2} – 3x + 6y + k = 0$
or ${x^2} + {y^2} – \frac{3}{2}x + 3y + \frac{k}{2} = 0$ …..$(i)$
and ${x^2} + {y^2} – 4x + 10y + 16 = 0$ …..$(ii)$
Circle $(i)$ and $(ii)$ cut orthogonally,
then $2{g_1}\,{g_2} + 2{f_1}{f_2} = {c_1} + {c_2}$
$2{\rm{ }}\left( { – \frac{3}{4}} \right)\,( – 2) + 2{\rm{ }}\left( {\frac{3}{2}} \right)\,.\,5 = \frac{k}{2} + 16$
$3 + 15 = \frac{k}{2} + 16$
==> $18 = \frac{k}{2} + 16$
==> $k = 4.$
Standard 11
Mathematics
