Question

Find the points on the circle x^2+y^2=80 which is nearest to (1,2)

Answer 1

Answer:

Please see the attachment

Answer 2

The points are (4,8) and (-4,-8).

equation of the circle - x² + y² = 80, center of the circle (0,0).

The point (1,2) will lie inside the given circle.

The point nearest to (1,2) lying on the circle will lie on the line from the center of the circle containing the point (1,2).

The line containing the points , (0,0) and (1,2) -

y = 2x

let the point on the circle be (h,k).

h = 2k, the point satisfies the equation .

h² + k² = 80 , point lies on the circle

=> (2k)² + k² = 80

=> 5k² = 80

=> k² = 16

=> k = 4,-4

h = 8,-8

The points that are possible - (4,8) and (-4,-8).