If a circle $S$ passing through the points $A(1,2)$ and $B(2,1)$ has its centre $\mathrm{C}$ located in the…

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 $\mathrm{P}(1,-2)$
  1. lies inside the circle $\mathrm{S}$
  2. lies outside the circle $\mathrm{S}$
  3. lies on the circle $\mathrm{S}$
  4. lies on the line $\mathrm{AB}$

Solution


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.

Asked in: AP EAMCET 2023 (16 May Shift 2)

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