A circle of radius 3 is centered at the origin and passes through the point (2,−5‾‾√).

(a) Find an equation for the line through the origin and the point (2,−5‾‾√).
y=

(b) Find an equation for the tangent line to the circle at (2,−5‾‾√).
y=

I expect you meant to say

(2,-√5)

man, some of the weird notations folks come up with!

The line passing through the origin with slope m is

y = mx

You have a point, so you can figure the slope, right?

The tangent line to a circle at any point (x,y) is -x/y

SO, you want to use the point-slope form for a line through (2,-√5) with slope 2/√5

y+√5 = 2/√5(x-2)

See the graphs at

http://www.wolframalpha.com/input/?i=plot+x^2%2By^2%3D9%2C+y+%3D+2%2F%E2%88%9A5%28x-2%29-%E2%88%9A5

To find an equation for a line, we can use the point-slope form of a linear equation:

y - y1 = m(x - x1),

where (x1, y1) is a point on the line, and m is the slope of the line.

(a) Finding the equation for the line passing through the origin (0, 0) and the point (2, -√5):

1. Find the slope (m):
m = (y2 - y1) / (x2 - x1) = (-√5 - 0) / (2 - 0) = -√5 / 2.

2. Substitute the values into the point-slope equation:
y - 0 = -√5/2(x - 0).

Simplifying the equation, we get:
y = -√5/2x.

Therefore, the equation for the line is y = -√5/2x.

(b) To find the equation for the tangent line to the circle at the point (2, -√5):

1. Determine the slope of the tangent line. Since the tangent line is perpendicular to the radius that intersects the point (2, -√5), the slope of the tangent line will be the negative reciprocal of the slope of the radius.

The radius that passes through (2, -√5) and the center of the circle (0, 0) has a slope of:
m = (y2 - y1) / (x2 - x1) = (-√5 - 0) / (2 - 0) = -√5 / 2.

The negative reciprocal of -√5 / 2 is 2 / √5.

2. Substitute the values into the point-slope equation using the point (2, -√5):
y - (-√5) = (2/ √5)(x - 2).

Simplifying the equation, we get:
y + √5 = (2/ √5)x - 4/√5.

To eliminate the square root in the denominator, we can multiply numerator and denominator by √5:
y + √5 = (2/ √5)x - 4/√5 * √5 / √5.

Simplifying further, we get:
y + √5 = (2√5 / 5)x - 4 / 5.

Finally, rearrange the equation to get it in the standard form:
y = (2√5 / 5)x - 4 / 5 - √5.

Therefore, the equation for the tangent line is:
y = (2√5 / 5)x - (4 + √5) / 5.