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If a circle $S$ passing through the points $A(1,2)$ and $B(2,1)$ has its centre $\mathrm{C}$ located in the third quadrant at a distance of $\frac{7}{\sqrt{2}}$ units from $\mathrm{AB}$, then the point $\mathrm{P}(1,-2)$
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Verified Answer
The correct answer is:
lies inside the circle $\mathrm{S}$
Now,
$\begin{aligned}
& \mathrm{AP}=\frac{1}{2} \sqrt{1^2+(-1)^2}=\frac{1}{\sqrt{2}} \\
& \text { and } \mathrm{PC}=\frac{7}{\sqrt{2}} \\
& \text { so } \mathrm{AC}=\sqrt{\left(\frac{1}{\sqrt{2}}\right)^2+\left(\frac{7}{\sqrt{2}}\right)^2}=5
\end{aligned}$
Now, $\mathrm{AC}=\mathrm{BC} \Rightarrow \mathrm{AC}^2=\mathrm{BC}^2$
$\begin{aligned}
& \Rightarrow(\mathrm{h}-1)^2+(\mathrm{k}-2)^2=(\mathrm{h}-2)^2+(\mathrm{k}-1)^2 \\
& \Rightarrow \mathrm{h}=\mathrm{k} ... (i)
\end{aligned}$
$\begin{aligned}
& \text { And } A C^2=5^2 \Rightarrow(h-1)^2+(h-2)^2=25 \\
& \Rightarrow h^2+1-2 h+h^2+4-4 h=25 \\
& \Rightarrow 2 h^2-6 h-20 \Rightarrow h^2-3 h-10=0 \\
& \Rightarrow(h-5)(h+2)=0 \Rightarrow h=-2,5
\end{aligned}$
So $(h, k)=(-2,-2)$
Distance between $(1-2) \&(-2,-2)=3$ and $3 < 5$
So the point $\mathrm{P}(1,-2)$ lies inside the circle.
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