Answer:
[tex]\qquad\boxed{ \sf{ \:\bf \: Coordinates \: of \: point \: = \bigg( \dfrac{1}{2}, \: \dfrac{1}{4} \bigg) \: }} \\ \\ [/tex]
Step-by-step explanation:
Given equation of circle is x² + y² = 1
Let assume that (h, k) be any point on the line 2x + y = 4.
[tex]\sf\implies \sf \:2h + k = 4 - - - (1) \\ \\ [/tex]
Now, Equation of chord of contact to the given circle is
[tex]\sf \:hx + ky = 1 \\ \\ [/tex]
can also be rewritten as
[tex]\sf\implies \sf \:4hx + 4ky = 4 - - - (2) \\ \\ [/tex]
On Subtracting equation (1) from equation (2), we get
[tex]\sf\implies \sf \:4hx + 4ky - 2h - k = 0 \\ \\ [/tex]
[tex] \sf \:4hx - 2h+ 4ky -k = 0 \\ \\ [/tex]
[tex] \sf \:(2x - 1)2h+ (4y - 1)k = 0 \\ \\ [/tex]
[tex]\sf\implies 2x - 1 = 0 \: \: \: and \: \: \: 4y - 1 = 0 \\ \\ [/tex]
[tex]\sf\implies x = \dfrac{1}{2} \: \: \: and \: \: \: y = \dfrac{1}{4} \\ \\ [/tex]
Hence,
[tex]\bf\implies Coordinates \: of \: point \: = \bigg( \dfrac{1}{2}, \: \dfrac{1}{4} \bigg) \\ \\ [/tex]
[tex]\rule{190pt}{2pt}[/tex]
Alternative Method
Let assume that (h, k) be the point on the line 2x + y = 0.
[tex]\sf\implies \sf \:2h + k = 4 - - - (1) \\ \\ [/tex]
Now, Equation of chord of contact is
[tex]\sf \:hx + ky = 1 \\ \\ [/tex]
[tex]\sf \:hx + (4 - 2h)y = 1 \\ \\ [/tex]
[tex]\sf \:hx + 4y - 2hy - 1 = 0\\ \\ [/tex]
[tex]\sf \:(4y - 1) + h(x - 2y) = 0\\ \\ [/tex]
It means, it passes through intersection of two lines
[tex]\sf \:4y - 1 = 0 \: \: and \: \: x - 2y = 0\\ \\ [/tex]
[tex]\sf\implies \sf \:y = \dfrac{1}{4} \: \: and \: \: x = 2y\\ \\ [/tex]
[tex]\sf\implies \sf \:y = \dfrac{1}{4} \: \: and \: \: x =\dfrac{1}{2} \\ \\ [/tex]
Hence,
[tex]\bf\implies Coordinates \: of \: point \: = \bigg( \dfrac{1}{2}, \: \dfrac{1}{4} \bigg) \\ \\ [/tex]
