- Home
- Standard 11
- Mathematics
10-1.Circle and System of Circles
hard
Give the number of common tangents to circle ${x^2} + {y^2} + 2x + 8y - 23 = 0$ and ${x^2} + {y^2} - 4x - 10y + 9 = 0$
A
$1$
B
$3$
C
$2$
D
None of these
Solution
(c) ${x^2} + {y^2} + 2x + 8y – 23 = 0$
$\therefore {C_1}( – 1, – 4),{r_1} = 2\sqrt {10} $
Again ${x^2} + {y^2} – 4x – 10y + 9 = 0$
${C_2}(2,5),{r_2} = 2\sqrt 5 $
Now ${C_1}{C_2}$ = distance between centres.
$\therefore $ ${C_1}{C_2} = \sqrt {9 + 81}$
$= 3\sqrt {10} = 9.486$ and
${r_1} + {r_2} = 2(\sqrt {10} + \sqrt 5 ) = 10.6$
${r_1} – {r_2} = 2\sqrt 5 (\sqrt 2 – 1) = 2 \times 2.2 \times 0.4 $
$= 4.4 \times 0.4 = 1.76$
${C_1}{C_2} = 2\sqrt {10} > {r_1} – {r_2}$
${r_1} – {r_2} < {C_1}{C_2} < {r_1} + {r_2} $
$\Rightarrow $ Two tangents can be drawn.
Standard 11
Mathematics
