Slope of line passing through origin and tangent to circle.

Question

If a line passing through origin touches the circle x² + y² - 6x + 8y + 9 = 0, then its slope may be :

(1)12/7

(2)-12/7

(3)24/7

(4)-24/7

Answer 1

We may rewrite the formula of the circle as x^2 - 6x + 9 - 9 + y^2 + 8y + 16 - 16 + 9 = 0, or

(x-3)^2 + (y+4)^2 = 16, a circle centered at (3, -4) with radius 4. Note that a line from the origin will be tangent to the circle at (3, 0), which is a slope of 0. The center of the circle has a slop of -4/3 from the origin. Thus, we can see that, via symmetry, using the tan 2a formula, there will be a tangent with 2 * -4/3/(1- (-4/3)^2) =

-8/3/(-7/9)) = -8/3 * -9/7 = 24/7

To verify, this tangent makes for a 7-24-25 right triangle.

The point 3 from the origin is (-21/25, -72/25)

This point is (-21/25 - 3, -72/25 + 4) = (-96/25, 28/25) = 4 from the center of the circle at a slope of -7/24 (correct distance and with a negative reciprocal slope implies perpendicular to the line from the origin).

Answer 2

1) Use completion of the square to put equation into graphing form.

2) Locate the point of tangency

3) Slope is negative reciprocal of the slope of the radius to the point of tangency.