Question

It is known that the straight line l is tangent to the circle x^2+y^2=4 at a point on the x-axis, and intersects with the straight line y=(1/2)x at the point P, find the coordinates of the point P​

Answer 1

Step-by-step explanation:

The general equation of a circle is

                                         $(x \ - \ h)^2 \ + \ (y \ - \ k)^2 \ = \ r^{2}$,

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

                        .                            $x^2 \ + \ y^2 \ = r^2$.

Now, we are interested in solving for the x-intercepts (the x-coordinates when the circle intersects the x-axis), of the circle

                                                      $x^2 \ + \ y^2 \ = \ 4$ .

Thus,

                                                     $x^2 \ + \ (0)^2 \ = \ 4 \\ \\ \\ \-\hspace{1.3cm} x^{2} \ = \ 4 \\ \\ \\ \-\hspace{1.4cm} x \ = \ \pm \ 2$.

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 $x \ = \ \pm \ 2$.

First and foremost, for the vertical line $x \ = \ 2$, it intersects the straight line $y \ = \ \displaystyle\frac{1}{2}x$  , giving the y-coordinate for point P,

                                                    $y \ = \ \displaystyle\frac{1}{2}(2) \\ \\ \\ y \ = \ 1$.

Hence, the coordinates of point P are $(1, \ 1)$.

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 $x \ = \ -2$ and interdects the straight line $y \ = \ \displaystyle\frac{1}{2}x$.

                                                      $y \ = \ \displaystyle\frac{1}{2}(-2) \\ \\ \\ y \ = \ -1$.

Therefore, the coordinates of point P are also $(1, \ -1)$.