a) When x = -2, in order to be on the circle, y must satisfy the equation
4 + y^2 = 40. That means
y^2 = 36.
y can be either 6 or -6. Therefore the point (-2,-6) IS on the circle.
b) The origin of the circle is (0,0). The equation of the circle tells you that. A line passing from there to the point (-2,-6) has slope (-6/-2) = 3. Its equation is y = 3x. (There is no constant term, since the y-intercept is 0)
c)The slope of a perpendicular line must be -1/3. It must pass through (-2,-6) go thorugh A. The equation of that line is:
[y-(-6)]/[x-(-2)] = -1/3
(y+6)/(x+2) = -1/3
y+6 = -x/3 -2/3
y = -x/3 -20/3
a)determine whether the point A(-2, -6) lies on the circle defined by x^2 + y^2 = 40
b) find an equation for the radius from the origin O to point A
c) Find an equation for the line that passes through A and is perpendicular to OA
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