which point on the circle x^2+y^2-12x-4y=50 is closest to the origin? Which point is farthest from the origin?

Question

Please help me find the answer.

Answer 1

Draw a diagram!

x² + y² - 12x - 4y = 50

(x - 6)² + (y - 2)² = 90         completion of square

 

Center (6, 2)

 

The two points shall be on a line passing thought the center and the origin and also intersecting the circle.

 

1) determine the equation of the line

2) determine the common solution to the line and the circle.

Answer 2

The idea here is to consider the line containing the origin and the center of the circle, and find the intersection of this line with the given circle. You will get two points, one which closest to the origin and the other which is farthest from the origin. 

 

Rewrite the equation of the circle in standard form by completing the square.

 

x² - 12x + 36 + y² - 4y + 4 = 50 + 36 + 4

 

(x - 6)² + (y - 2)² = 90

 

The center of the circle (6,2), so the line containing (6,2) and (0,0) is y = (1/3)x. The intersection of the line and circle are found by solving the equations simultaneously. Substitute (1/3)x for y in the equation x² + y² - 12x - 4y = 50 to get

 

x² + (x/3)² - 12x - 4x/3 = 50

 

x² + x²/9 - 12x - 4x/3 = 50

 

10x²/9 - 40x/3 - 50 = 0

 

x² - 12x - 45 = 0  [multiply the previous equation by 9/10]

 

(x - 15)(x + 3) = 0

 

x = 15, -3

 

y = 5, -1

 

The points of intersection are (15, 5) and (-3, -1). The point closest to the origin is (-3, -1) and (15, 5) is the point farthest from the origin.