Centre of circle passing through (0, 0) and (1, 0) and touching circle {x^2} + {y^2} = 9 is

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10-1.Circle and System of Circles

hard

The centre of the circle passing through $(0, 0)$ and $(1, 0)$ and touching the circle ${x^2} + {y^2} = 9$ is

$\left( {\frac{1}{2},\frac{1}{2}} \right)$

$\left( {\frac{1}{2}, - \sqrt 2 } \right)$

$\left( {\frac{3}{2},\frac{1}{2}} \right)$

$\left( {\frac{1}{2},\frac{3}{2}} \right)$

(AIEEE-2002)

Solution

(b) If two circles touch and one of them passes through the centre $(0,0)$ of the other then they must touch internally.

$\therefore $ ${C_1}{C_2} = {r_1} – {r_2}$

Also, ${C_1}{C_2} = {r_2}$

$\therefore $ ${r_2} = {r_1} – {r_2}$

$ \Rightarrow $ ${r_2} = \frac{{{r_1}}}{2} = \frac{3}{2}$

Now, suppose the equation of circle ${S_2}$ is ${x^2} + {y^2} + 2gx + 2fy + c = 0$

$\therefore $ It passes through $(0, 0)$, $\therefore $ $c = 0$

Also, It passes through $(1, 0)$

$\therefore $ $2g + 1 = 0$

$ \Rightarrow \,\,\,g = – \frac{1}{2}$

Hence ${C_2}\left( {\frac{1}{2},\, – f} \right)$.

Now $\sqrt {{{\left( {\frac{1}{2}} \right)}^2} + {f^2}} = {r_2}$

$ \Rightarrow $ $\frac{1}{4} + {f^2} = {\left( {\frac{3}{2}} \right)^2}$

$ \Rightarrow $ $f = \, \pm \,\,\sqrt 2 $

Hence the centre of required circle is $\left( {\frac{1}{2},\, – \sqrt 2 } \right)$.

Standard 11

Mathematics

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The centre of the circle passing through $(0, 0)$ and $(1, 0)$ and touching the circle ${x^2} + {y^2} = 9$ is

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