1 \(\left(\frac{12}{25}, \frac{36}{25}\right)\)
2 \((4,0)\)
3 \((2,0)\)
4 \(\left(\frac{-14}{25}, \frac{48}{25}\right)\)
Explanation:
C Let the point be (h, \(\mathrm{k}\) )
So,
\(\frac{|4 \mathrm{~h}+3 \mathrm{k}-12|}{5}=\frac{4}{5} \Rightarrow|4 \mathrm{~h}+3 \mathrm{k}-12|=4\)
\(\Rightarrow \quad(4 \mathrm{~h}+3 \mathrm{k}=16)\) or \((4 \mathrm{~h}+3 \mathrm{k}=8)\)
\((\mathrm{h}, \mathrm{k})\) lies on circle so
\(\quad \mathrm{h}^2+\mathrm{k}^2=4\)
\(\Rightarrow\left[\mathrm{h}^2+\left(\frac{16-4 \mathrm{~h}}{3}\right)^2=4\right] \text { or }\left[\mathrm{h}^2+\left(\frac{8-4 \mathrm{~h}}{3}\right)^2=4\right]\)
\(\Rightarrow\left(25 \mathrm{~h}^2-128 \mathrm{~h}+220=0\right) \text { or }\left(25 \mathrm{~h}^2-64 \mathrm{~h}+28=0\right)\)
Since
\(25 h^2-128 h+220=0\)
has no real roots we have
\(25 h^2-64 h+28=0\)
and hence \(\left(\mathrm{h}=2, \frac{14}{25}\right)\)
At \(\mathrm{h}=2, \mathrm{k}=0\)
At \(\mathrm{h}=\frac{14}{25}, \mathrm{k}=\frac{48}{25}\)
So, \((\mathrm{h}, \mathrm{k})=(2,0)\) or \(\left(\frac{14}{25}, \frac{48}{25}\right)\)