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10-1.Circle and System of Circles
medium
The value of k so that ${x^2} + {y^2} + kx + 4y + 2 = 0$ and $2({x^2} + {y^2}) - 4x - 3y + k = 0$ cut orthogonally is
A
$\frac{{10}}{3}$
B
$\frac{{ - 8}}{3}$
C
$\frac{{ - 10}}{3}$
D
$\frac{8}{3}$
Solution
(c) Here, ${g_1} = \frac{k}{2},\,{f_1} = 2,\,{c_1} = 2$
${g_2} = – 1,\,{f_2} = \frac{{ – 3}}{4},\,{c_2} = \frac{k}{2}$
Condition for orthogonal intersection,
==> $2({g_1}{g_2} + {f_1}{f_2}) = {c_1} + {c_2}$
==> $2\,\left[ {\frac{{ – k}}{2} + \left( {\frac{{ – 3}}{2}} \right)} \right] = 2 + \frac{k}{2}$
==> $ – k – 3 = 2 + \frac{k}{2}$
==> $\frac{{3k}}{2} = – 5$;
$k = \frac{{ – 10}}{3}$.
Standard 11
Mathematics
