Give the number of common tangents to circle | vedclass

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(c) ${x^2} + {y^2} + 2x + 8y - 23 = 0$

$	herefore {C_1}( - 1, - 4),{r_1} = 2\sqrt {10} $

Again ${x^2} + {y^2} - 4x - 10y + 9 = 0$ 

${C_

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$2$

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(c) ${x^2} + {y^2} + 2x + 8y - 23 = 0$

$	herefore {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.

$	herefore $ ${C_1}{C_2} = \sqrt {9 + 81}$

$= 3\sqrt {10} = 9.486$ and

${r_1} + {r_2} = 2(\sqrt {10} + \sqrt 5 ) = 10.6$

${r_1} - {r_2} = 2\sqrt 5 (\sqrt 2 - 1) = 2 	imes 2.2 	imes 0.4 $

$= 4.4 	imes 0.4 = 1.76$

${C_1}{C_2} = 2\sqrt {10} > {r_1} - {r_2}$

${r_1} - {r_2} < {C_1}{C_2} < {r_1} + {r_2} $

$\Rightarrow $ Two tangents can be drawn.

Give the number of common tangents to circle ${x^2} + {y^2} + 2x + 8y - 23 = 0$ and ${x^2} + {y^2} - 4x - 10y + 9 = 0$

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