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A circle passes through the points $(2,3)$ and $(4$, 5). If its centre lies on the line, $y-4 x+3=0$, then its radius is equal to
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Verified Answer
The correct answer is:
$\sqrt{2}$
$\sqrt{2}$
Equation of the line passing through the points $(2,3)$ and $(4,5)$ is
$$
y-3=\left(\frac{5-3}{4-2}\right) x-2 \Rightarrow x-y+1=0
$$
Equation of the perpendicular line passing through the midpoint $(3,4)$ is $x+y-7=0$
Lines (1) and (2) intersect at the center of the circle. So, the center of the circle is $(3,4)$ Therefore, the radius is
$$
\begin{aligned}
&\sqrt{\left(x_2-x_1\right)^2+\left(y_2-y_1\right)^2} \\
&=\sqrt{(2-3)^2+(3-4)^2}=\sqrt{2} \text { units. }
\end{aligned}
$$
$$
y-3=\left(\frac{5-3}{4-2}\right) x-2 \Rightarrow x-y+1=0
$$
Equation of the perpendicular line passing through the midpoint $(3,4)$ is $x+y-7=0$
Lines (1) and (2) intersect at the center of the circle. So, the center of the circle is $(3,4)$ Therefore, the radius is
$$
\begin{aligned}
&\sqrt{\left(x_2-x_1\right)^2+\left(y_2-y_1\right)^2} \\
&=\sqrt{(2-3)^2+(3-4)^2}=\sqrt{2} \text { units. }
\end{aligned}
$$
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