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If one of the diameters of the curve $x^{2}+y^{2}-4 x-6 y+9=0$ is a chord of a circle with centre $(1,1),$ the radius of this circle is
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3
Given circle
$x^{2}+y^{2}-4 x-6 y+9=0$
$\therefore$ Centre $=\left(-\frac{1}{2} \times(-4),-\frac{1}{2} \times(-6)\right)$
$=(2.3)$
$\begin{aligned} \text { Radius } &=\sqrt{(-2)^{2}+(-3)^{2}-9} \\ &=\sqrt{4+9-9} \\ &=2 \end{aligned}$
$$
\begin{aligned}
\therefore \quad A B &=\sqrt{(2-1)^{2}+(3-1)^{2}}=\sqrt{1+4}=\sqrt{5} \\
\therefore \quad A C &=\sqrt{A B^{2}+B C^{2}}=\sqrt{(\sqrt{5})^{2}+(2)^{2}} \\
&=\sqrt{5+4}=\sqrt{9}=3
\end{aligned}
$$
Radius of required circle $=3$
$x^{2}+y^{2}-4 x-6 y+9=0$
$\therefore$ Centre $=\left(-\frac{1}{2} \times(-4),-\frac{1}{2} \times(-6)\right)$
$=(2.3)$
$\begin{aligned} \text { Radius } &=\sqrt{(-2)^{2}+(-3)^{2}-9} \\ &=\sqrt{4+9-9} \\ &=2 \end{aligned}$
$$
\begin{aligned}
\therefore \quad A B &=\sqrt{(2-1)^{2}+(3-1)^{2}}=\sqrt{1+4}=\sqrt{5} \\
\therefore \quad A C &=\sqrt{A B^{2}+B C^{2}}=\sqrt{(\sqrt{5})^{2}+(2)^{2}} \\
&=\sqrt{5+4}=\sqrt{9}=3
\end{aligned}
$$
Radius of required circle $=3$
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