Solving linear quadratic systems

Question

A fenced-in, square park is 30 feet on each side. The entrance to the park is in the center of the south edge of the park. At the center of the park is a fountain with a diameter of 10 feet Luis is meeting Samantha in the park. Samantha is 25 feet north and 5 feet west of the entrance. Can Luis walk a straight line path from the entranice to where Samantha is sitting? Explain your answer.

Answer 1

If you construct a diagram with the fountain midpoint at (0,0) of a coordinate system, the the equation of the circle is x² + y² = 25.

There was no information on the width of the gate so I assumed that the person had to start exactly at point (0,-15) the exact center of the gate. Based on the information, Samantha would be sitting at point (-5,10).

The line from the gate to Samantha is y = -5x -15. So if you set the graphs of the circle and the line equal to one another, you obtain a quadratic equation where the x values that satisfy this condition are x = -3.6775 and -2.0917.

The question is...is this really an intersection with the circle & line?

Line Circle

x= -3.6775

y= 3.3875 y = +- 3.3876

x= -2.0917

y= -4.4515 y = +- 4.5414

It does appear that the line & circle do coincide at 2 points so a straight line is not possible!