MATH 216T Homework 1 Solutions 1. Find the greatest common ...

( c)(We easily get that the X-intercept of ax+by = c isa ,0 while the Y-intercept is 0, c ). HencebP(0, c b )❩ ❩❩❩❩❩❩❩❩❩❩❩Rr✓ ✓✓✓✓✓O(0,0)Q( c a ,0)Using similar triangles, we get that |OR||OQ| = |OP||PQ| . Hencer = |OR| = |OQ|·|OP||PQ|=√ (caca · cb) 2 (+c)= ··· =2bc√a2 +b 2Hence the line ax+by = c will intersect the circle x 2 +y 2 = r 2 in two different points if and only if r >c√a2 +b 2.Note that we can get this result pure algebraically. There are two points of intersection if and only if the system{ ax+by = c (1)x 2 +y 2 = r 2 (2)has two solutions. It follows from (1) that y = c−ax . Substituting this into (2), we findb( ) 2 c−axx 2 + = r 2bor(a 2 +b 2 )x 2 −2acx+c 2 −b 2 r 2 = 0This quadratic equation is x has two solutions if and only if it’s discriminant is strictly positive :(−2ac) 2 −4(a 2 +b 2 )(c 2 −b 2 r 2 ) > 0Solving for r 2 , we get r 2 > c2a 2 +b 2.√c2(b) Put r =a 2 +b 2 + a2 +b 2. Since r >4x 2 +y 2 = r 2 in two different points, say P 1 (x 1 ,y 1 ) and P 2 (x 2 ,y 2 ). Suppose we can prove that the distance d betweenP 1 and P 2 is at least √ a 2 +b 2 . Then it follows from what we have seen in class that there exists a couple (x 0 ,y 0 ) ∈ Z 2such that ax 0 + by 0 = c and the point P(x 0 ,y 0 ) belongs to the line segment P 1 P 2 . Hence P lies inside the circlex 2 +y 2 = r 2 . Soc√ it follows from (a) that the ax+by = c will intersect the circlea2 +b2, x 2 0 +y0 2 ≤ r 2 = c2a 2 +b 2 + a2 +b 24It remains to prove that d ≥ √ a 2 +b 2 . This is equivalent to showing that d 2 ≥ a 2 +b 2 .We have the following picture :4