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Q.
Two circles of equal radius $'a'$ cut orthogonally. If their centres are $(2, 3)$ and $(5, 6)$, then radical axis of these circles passes through the point
TS EAMCET 2017
Solution:
Let $S_{1}$ be the circle with centre $(2,3)$ and radius a.
Let $S_{2}$ be the circle with centre $(5,6)$ and radius $a$.
Then $S_{1} \equiv(x-2)^{2}+(y-3)^{2}-a^{2}=0$
and $S_{2}=(x-5)^{2}+(y-6)^{2}-a^{2}=0$
Now, radical axis of these circle is given by $S_{1}-S_{2}=0$
$(x-2)^{2}+(y-3)^{2}-a^{2}-(x-5)^{2}-(y-6)^{2}+a^{2}=0$
$\Rightarrow (4-4 \,x+9-6\, y)-(25-10\, x)-(36-12 \,y)=0$
$\Rightarrow 13-4 \,x-6 \,y-25+10 x-36+12\, y=0$
$\Rightarrow 6\, x+6\,y-48=0$
$\Rightarrow x+y=\frac{48}{6}=8$
$\Rightarrow x+y=8\,....(i)$
Since, the given circle cut orthogonally, therefore
$2 g_{1} g_{2}+2 f_{1} f_{2}=c_{1}+c_{2}$
$ \Rightarrow 2(-2)(-5)+2(-3)(-6)=\left((-2)^{2}+(-3)^{2}-a^{2}\right) $
$+\left((-5)^{2}+(-6)^{2}-a^{2}\right) $
$[\because \text { radius }=\sqrt{g^{2}+f^{2}-c} \therefore a^{2}=g^{2}+f^{2}-c \Rightarrow .$
$ \left.C=g^{2}+f^{2}-a^{2}\right] $
$ \Rightarrow 20+36=(13-a^{2})+(61-a^{2}) $
$ \Rightarrow 56=74-2 a^{2}.$
$\Rightarrow 2 a^{2}=74-56=18$
$\Rightarrow a^{2}=9$
$\Rightarrow a=3[\because$ radius can't be negative $]$
Clearly, option (c) satisfy the Eq. (i)