Find the point(s) where the line through the origin with slope 6 intersects the unit circle.
The line passes through the origin so its equation is y=6x. The equation of the unit circle is x^2+y^2=1, which has centre (0,0) and radius 1. Substituting y=6x in the equation of the circle we have 37x^2=1 and x=+sqrt(1/37). Therefore the y values for the intersections are y=+6sqrt(1/37). The points of intersection are (-sqrt(1/37),-6sqrt(1/37)), (sqrt(1/37),6sqrt(1/37)).
If the unit circle has centre (h,k) the equation is (x-h)^2+(y-k)^2=1 and substituting y=6x we get (x-h)^2+(6x-k)^2=1, which gives the x value of the intersection. So x^2-2xh+h^2+36x^2-12xk+k^2=1; 37x^2-2x(h+6k)+h^2+k^2-1=0. There are only two factors of 37, which is prime, so to factorise rationally we must have (37x+a)(x+b)=0; 37x^2+x(37b+a)+ab=0. Using the quadratic formula: x=(h+6k+sqrt((h+6k)^2-37(h^2+k^2-1))/37. The square root can only be evaluated if the expression is positive, so (h+6k)^2>37(h^2+k^2-1). This requirement applies so that the line intersects the unit circle. When the expression is zero, the line is a tangent to the circle, so there is only one intersection point. h^2+36k^2+12hk>37h^2+37k^2-37; 36h^2-12hk+k^2<37; (6h-k)^2<37 and (6h-k)< +sqrt(37) (=+6.08). This connects the coordinates of the centre of the unit circle: k>6h+sqrt(37). On equality the line y=6x will be tangential to the circle. For example, if h=0 (centre of the circle is on the y axis), k=+sqrt(37) and the unit circle will lie above or below the axis with y=6x running tangentially on the right of the circle; or on the left touching the circle below the x axis.
Ideally, we want the square root to be rational so 37-(6h-k)^2=a^2. If a=+1, 6h-k=6 and k=6(h-1); or 6h-k=-6, so k=6(h+1). That gives many possible values for h and k represented by pairs: (1,0), (2,6), (3,12), (0,-6), (-1,-12), (-2,-18), (0,6), (1,12), (-1,0), (-2,-6), to mention but a few. Using (1,0) in the quadratic: 37x^2-2x(h+6k)+h^2+k^2-1=0 we have 37x^2-2x=0=x(37x-2) giving intersection points (0,0) and (2/37,12/37). The equation for the circle is (x-1)^2+y^2=1 or y^2=2x-x^2.
Let's try (2,6). 37x^2-76x+39=0, (37x-39)(x-1)=0 giving intersection points (39/37,234/37) and (1,6). The equation of the circle is (x-2)^2+(y-6)^2=1 or x^2-4x+y^2-12x+39=0.
If a=+6, 6h-k=1 or -1, so k=6h-1 or 6h+1. This generates more possible intersection points. There are clearly an infinite number of positions for the unit circle centre (h,k) and an infinite number of intersection points. However, the relationship between h and k so as to produce rational intersection points has been established. k=6(h+1), k=6h+1 are the equations linking the coordinates of the centre of the unit circle. With these equations in mind the quadratic determining the intersection points (x,6x) can be solved: 37x^2-2x(h+6k)+h^2+k^2-1=0. There are four variations of this quadratic because there are four equations linking h and k.
There are 2 values of a^2 where a^2=37-(6h-k)^2 and x=(h+6k+a)/37; a^2=1 or 36.
When a^2=1, k=6(h+1). The equation of the circle is (x-h)^2+(y-6(h+1))^2=1 and x=(37h+36+1)/37. So the points of intersection are (h+1,6(h+1)), ((37h+35)/37,6(37h+35)/37), ((37h-35)/37,6(37h-35)/37), (h-1,6(h-1)).
When a^2=36, k=6h+1. The equation of the circle is (x-h)^2+(y-6h+1)^2=1 and x=(37h+6+6)/37. The points of intersection are ((37h+12)/37,6(37h+12)/37), (h,6h), ((37h-12)/37,6(37h-12)/37). Note that (h,6h) is the result of (37h+6-6)/37 and (37h-6+6)/37.
We can check the (h,k) values we used earlier. These were (1,0) and (2,6). We used the formula k=6(h-1) in each case (a=+1), so intersection points for h=1, k=6(h-1)=0, should be x=(h+6k+1)/37, giving (2/37,12/37) and x=(h+6k-1), giving (0,0). For (2,6) h=2 and k=6, giving intersection points x=(2+36+1)/37, giving (39/37,234/37) and x=(2+36-1)/37=1, giving (1,6).
The values of h and k are not restricted to integers. Read More: ...