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10-1.Circle and System of Circles
hard
The equation of the circle having the lines ${x^2} + 2xy + 3x + 6y = 0$ as its normals and having size just sufficient to contain the circle $x(x - 4) + y(y - 3) = 0$is
A
${x^2} + {y^2} + 3x - 6y - 40 = 0$
B
${x^2} + {y^2} + 6x - 3y - 45 = 0$
C
${x^2} + {y^2} + 8x + 4y - 20 = 0$
D
${x^2} + {y^2} + 4x + 8y + 20 = 0$
Solution
(b) Given circle is $\left( {2,\;\frac{3}{2}} \right){\rm{ }},\;\frac{5}{2} = {r_1}$ (say)
Required normals of circlres are $x + 3 = 0,\;x + 2y = 0$
which intersect at the centre $\left( { – 3,\;\frac{3}{2}} \right){\rm{ }},\;{r_2} = $ radius (say).
$2^{nd}$ circle just contains the $1^{st}$
$i.e.$, ${C_2}{C_1} = {r_2} – {r_1}$
$\Rightarrow {r_2} = \frac{{15}}{2}$.
Standard 11
Mathematics
