1996 AHSME Problems/Problem 25
Problem
Given that 
, what is the largest possible value that 
can have?
Solution
The first equation is a circle, so we find its center and radius by completing the square:
, so ![\[(x-7)^2 + (y-3)^2 = (x- 14x + 49) + (y^2 - 6y + 9) = 6 + 49 + 9 = 64.\]](http://latex.artofproblemsolving.com/f/f/e/ffe0e4fae2297f35b74b0b79bdc3057cdaa6e31c.png)
So we have a circle centered at 
with radius 
, and we want to find the max of .
The set of lines 
are all parallel, with slope 
. Increasing 
shifts the lines up and/or to the right.
We want to shift this line up high enough that it's tangent to the circle, but not so high that it misses the circle altogether. This means will be tangent to the circle.
Imagine that this line hits the circle at point 
. The slope of the radius connecting the center of the circle, , to tangent point will be 
, since the radius is perpendicular to the tangent line.
So we have a point, , and a slope of that represents the slope of the radius to the tangent point. Let's start at the point . If we go 
units up and 
units right from , we would arrive at a point that's 
units away. But in reality we want 
to reach the tangent point, since the radius of the circle is .
Thus, 
, and we want to travel 
up and 
over from the point to reach our maximum. This means the maximum value of occurs at 
, which is 
Plug in those values for 
and 
, and you get the maximum value of 
, which is option 
See also
| 1996 AHSME (Problems • Answer Key • Resources) | ||
| Preceded by Problem 24 |
Followed by Problem 26 | |
| 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 • 26 • 27 • 28 • 29 • 30 | ||
| All AHSME Problems and Solutions | ||
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