Problem:
Find the largest value of xy for pairs of real numbers (x,y) which satisfy (x−3)2+(y−3)2=6.
Answer Choices:
A. 3+22
B. 2+3
C. 33
D. 6
E. 6+23
Solution:
Let P=(x,y),A=(0,0),C=(3,3) and B be any point on the positive x-axis. The locus of P is the circle with center C and radius 6, and xy is the slope of segment AP. Clearly this slope is greatest when AP is tangent to the circle on the left side, as in the adjoining. figure (Note: 6<3). Let α=∠CAP. Since 4BAC=45∘; the answer is tan(α+45∘)=1−tanαtanα+1. Since ∠APC=90∘,tanα=PAPC. By the Pythagorean Theorem, PA=(AC)2−(PC)2=23. Thus tanα=21 and the answer is 3+22.
The maximum value of xy is the maximum value of the slope of a line which contains the origin and which intersects the circle (x−3)2+(y−3)2=6. Clearly the line of maximum slope, m, is tangent to the circle and is the steeper of the two lines through the origin and tangent to the circle. In short, it is the line containing AP in the figure on page 9. The following statements are equivalent:
y=mx is tangent to the circle;
the system of equations
y=mx,(x−3)2+(y−3)2=6,
has only one solution (x,y);
the quadratic equation (x−3)2+(mx−3)2=6, or
(m2+1)x2−6(m+1)x+12=0
has a double root;
the discriminant is zero, i.e.,
36(m+1)2−48(m2+1)=0
Solving the last equation for its larger root, one obtains m=3+22.
Note: this method is more general than the previous one. The circle may be replaced by any conic and the method will still work.
