Step-by-step explanation:
The general equation of a circle is
[tex](x \ - \ h)^2 \ + \ (y \ - \ k)^2 \ = \ r^{2}[/tex],
where h and k forms the coordinates of the centre of the circle.
When the circle has a centre at the origin, the equation reduces into
. [tex]x^2 \ + \ y^2 \ = r^2[/tex].
Now, we are interested in solving for the x-intercepts (the x-coordinates when the circle intersects the x-axis), of the circle
[tex]x^2 \ + \ y^2 \ = \ 4[/tex] .
Thus,
[tex]x^2 \ + \ (0)^2 \ = \ 4 \\ \\ \\ \-\hspace{1.3cm} x^{2} \ = \ 4 \\ \\ \\ \-\hspace{1.4cm} x \ = \ \pm \ 2[/tex].
Geometrically speaking, the tangent to the circle at the point defined by one of the x-intercepts of the circle is actually a vertical line, more specifically the lines [tex]x \ = \ \pm \ 2[/tex].
First and foremost, for the vertical line [tex]x \ = \ 2[/tex], it intersects the straight line [tex]y \ = \ \displaystyle\frac{1}{2}x[/tex] , giving the y-coordinate for point P,
[tex]y \ = \ \displaystyle\frac{1}{2}(2) \\ \\ \\ y \ = \ 1[/tex].
Hence, the coordinates of point P are [tex](1, \ 1)[/tex].
However, since there are no boundaries given in the question and a circle is symmetrical about its centre, thus, point P also exists when the vertical line [tex]x \ = \ -2[/tex] and interdects the straight line [tex]y \ = \ \displaystyle\frac{1}{2}x[/tex].
[tex]y \ = \ \displaystyle\frac{1}{2}(-2) \\ \\ \\ y \ = \ -1[/tex].
Therefore, the coordinates of point P are also [tex](1, \ -1)[/tex].
