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10-1.Circle and System of Circles
hard
The equation of the circle which touches the circle ${x^2} + {y^2} - 6x + 6y + 17 = 0$ externally and to which the lines ${x^2} - 3xy - 3x + 9y = 0$ are normals, is
A
${x^2} + {y^2} - 6x - 2y - 1 = 0$
B
${x^2} + {y^2} + 6x + 2y + 1 = 0$
C
${x^2} + {y^2} - 6x - 6y + 1 = 0$
D
${x^2} + {y^2} - 6x - 2y + 1 = 0$
Solution
(d) The given normals are $x – 3y = 0,\;x – 3 = 0$ which intersect at centre whose co-ordinates are $(3, 1)$.
The given circle is ${C_1}(3,\; – 3)$ $\;{r_1} = 1,$
${C_2}$ is $(3, 1)$ and ${r_2} = (?)$.
If the two circles touch externally, then ${C_1}{C_2} = {r_1} + {r_2} $
$\Rightarrow 4 = 1 + {r_2}$
$\Rightarrow {r_2} = 3$
$\therefore $${(x – 3)^2} + {(y – 1)^2} = {3^2}$ or ${x^2} + {y^2} – 6x – 2y + 1 = 0$.
Standard 11
Mathematics
