The point on the circle \(x^2+y^2=4\) whose distance from the line \(4 x+3 y-12=0\) is \(4 / 5\) units is…
The point on the circle \(x^2+y^2=4\) whose distance from the line \(4 x+3 y-12=0\) is \(4 / 5\) units is equal to
\(\left(\frac{12}{25}, \frac{36}{25}\right)\)
\((4,0)\)
\((2,0)\)
\(\left(\frac{-14}{25}, \frac{48}{25}\right)\)
Solution
Let the point be \((h, k)\)
So, \(\quad \frac{|4 h+3 k-12|}{5}=\frac{4}{5} \Rightarrow|4 h+3 k-12|=4\)
\(\Rightarrow \quad(4 h+3 k=16)\) or \((4 h+3 k=8)\)
\((h, k)\) lies on circle so
\(\begin{aligned}
& h^2+k^2=4 \\
& \Rightarrow \quad {\left[h^2+\left(\frac{16-4 h}{3}\right)^2=4\right] \text { or }\left[h^2+\left(\frac{8-4 h}{3}\right)^2=4\right] } \\
& \Rightarrow\left(25 h^2-128 h+220=0\right) \text { or }\left(25 h^2-64 h+28=0\right) \\
& \Rightarrow 25 h^2-128 h+220=0 \Rightarrow \text { Imaginary point }
\end{aligned}\)
So, \(\quad 25 h^2-64 h+28=0\) will be considered and \(\left(h=2, \frac{14}{25}\right)\)
At \(h=2, k=0\)
At \(h=\frac{14}{25}, k=\frac{48}{25}\)
So, \((h, k)=(2,0)\) or \(\left(\frac{14}{25}, \frac{48}{25}\right)\)