The centre of the circle passing through (0, | vedclass

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Question

(b) If two circles touch and one of them passes through the centre $(0,0)$ of the other then they must touch internally.

$	herefore $ ${C_1}{C_2}

Vedclass · Accepted Answer

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

Vedclass · Answer

(b) If two circles touch and one of them passes through the centre $(0,0)$ of the other then they must touch internally.

$	herefore $ ${C_1}{C_2} = {r_1} - {r_2}$

Also, ${C_1}{C_2} = {r_2}$

$	herefore $ ${r_2} = {r_1} - {r_2}$

$ \Rightarrow $ ${r_2} = \frac{{{r_1}}}{2} = \frac{3}{2}$

Now, suppose the equation of circle ${S_2}$ is ${x^2} + {y^2} + 2gx + 2fy + c = 0$

$	herefore $ It passes through $(0, 0)$, $	herefore $ $c = 0$

Also, It passes through $(1, 0)$

$	herefore $ $2g + 1 = 0$

$ \Rightarrow \,\,\,g = - \frac{1}{2}$

Hence ${C_2}\left( {\frac{1}{2},\, - f} ight)$.

Now $\sqrt {{{\left( {\frac{1}{2}} ight)}^2} + {f^2}} = {r_2}$

$ \Rightarrow $ $\frac{1}{4} + {f^2} = {\left( {\frac{3}{2}} ight)^2}$

$ \Rightarrow $ $f = \, \pm \,\,\sqrt 2 $

Hence the centre of required circle is $\left( {\frac{1}{2},\, - \sqrt 2 } ight)$.

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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