Circle Question

Question

Recall the equation for a circle with center (h,k) and radius r. At what point in the first quadrant does the line with equation y = 1 x + 4 intersect the circle with radius 4 and center (0, 4)?

 

 

Answer 1

Just by inspection of the two equations, one could see that the straight line passes through the center of the circle, which is the y-intercept of the equation of the straight line, and also becomes the radius of the circle going from x = 0 to some point of intersection in the first quadrant. Also, the slope of the line being 1 makes the angle being 45 degrees with respect to the x-axis. Hence,

 

x = r * cos 45 = 4 * cos 45 = 2 * sqrt (2) utilizing trigonometry

 

or

 

4 = x * sqrt (2). Solve for x in which gives x = 2 *sqrt (2) using the property of 45 degree 45 degree right triangles in which its hypotenuse is the product of a side and sqrt(2).

 

Now substitute x in the equation of the straight line to get the y coordinate. Hence,

 

y = 2 * sqrt (2) + 4

 

Therefore, the point of intersection is (2 sqrt (2), 2 * sqrt (2) + 4)

 

 

Answer 2

Equation of circle

 

x^2+(y-4)^2=16

 

Equation of the line

 

y =x+4

 

Substituting in the equation of circle

 

x^2+x^2=16

 

2x^2=16

 

x^2=8

 

x=2*sqrt(2)

 

y=2*sqrt(2)+4

 

This is the point at which the circle intersects the line.