Write an equation for a circle entered at the orgin that passes through the points (-8,6)

Question

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Answer 1

The equation for a circle is (x - a)² + (y - b)² = r² where (a, b) is the center and r is the radius.

 

In this case the center is the origin, (0, 0), so that makes things easier:  x² + y² = r²

 

So all we have to do is find the radius.  The radius is the distance from the center to any point on the circle.  Since (-8, 6) is on the circle, the radius is just the distance between (-8, 6) and (0, 0).  Use the distance formula to determine that:

 

r = √((-8 - 0)² + (6 - 0)²) = √((-8)² + (6)²) = √(64 + 36) = √100 = 10

 

So r² = 100, which leaves you with the equation:  x² + y² = 100

Answer 2

The general equation of a circle centered at the origin is

 

x² + y² = r²

 

where r is the radius. 

 

Plug in the given coordinates to find the value of r².

 

(-8)² + (6)² = r²

 

64 + 36 = r²

 

100 = r²

 

 

The equation for a circle that passes through the point (-8,6) is

 

x² + y² = 100

Answer 3

We need to compute the radius of the circle...which would be given geometrically by the distance from the origin to the point (8,6).

 

distance from origin to (x,y) = dist(x,y) = \sqrt{x^2+y^2} (the euclidean 2-norm)

 

To see that this is true consider the Pythagorean theorem on the triangle form by dropping a perpendicular down to the x-axis from the pt (x,y) in the plane.

 

So the distance from (8,6) to the origin is given by \sqrt(8^2+6^2} = \sqrt(100) = 10...this is the radius of the circle we seek that is centered at the origin...and since circles centered at the origin are given by the formula x^2+y^2=r^2, we have:

 

x^2+y^2 = 100 as the equation for the circle centered at the origin passing through the point (8,6).