Circle Problem. SAT?

Question

Found it in my homework packet. I used calculus, but I feel like using pre-cal this time =D!

So here's the problem:

"In figure above, the circle with center C and radius 5 is tangent to both the x and y axes. Line AB is tangent to the circle at point B. What is the slope of Line AB if point B is at (9, -8)?

The circle is at the 4th quadrant. You might want to draw it to do this.

How would you find the slope without calculus?

You might want the equation of the circle:

(x-5)^2 + (y+5)^2 = 25

Thanks!

And I actually did my homework:

(x-5)^2 + (y+5)^2 = 25
2(x-5) + 2(y+5)(y') = 0
2(y+5)(y') = -2(x-5)
y' = -2(x-5) / 2(y+5)
y' = -(x-5) / (y+5)
(x,y) = (9,-8)
y' = -(9-5) / (-8+5)
y' = 4/3

I want to somehow get 4/3 without calculus.

Peter.... That's exactly what I did....

Please don't use formulas lol.

I see... sorry for thinking you were using calculus guys... Thanks! (excluding the first answerer though =)!)

Answer 1

Hi,

The slope of the radius from the circle center at (5,-5) to the point of tangency (9,-8) is m = ((-8) - (-5))/(9 - 5) = -3/4.

Since the radius and tangent line are always perpendicular, you can find the slope of the tangent line by finding the negative reciprocal of the slope of the radius. -(the reciprocal of -3/4) = 4/3. That's the slope of the tangent line, without calculus.

I hope that helps!! :-)

Answer 2

A line through the center of the circle, (5,-5) and the point
(9,-8) has slope = - 3/4
This line is perpendicular to the line AB since the radius is
perpendicular to the tangent line.
If a line with slope m is perpendicular to a line with slope M, then M = - 1/m
Thus the slope of line AB is - 1 / -3/4 = 4/3

Answer 3

Geometry is really all you need.
If the circle is tangent to both the x & y axis the center is at (5,-5). Sketch the circle with the tangent piont. The tangent is perpendicular to the radius. The slope from (5,-5) to (9,-8) is (-5 +8)/(5 -9) = 3/(-4) so the slop of the tangent line is 4/3

Answer 4

Slope of line joining centre (5, -5) and B(9, -8)
m = (-8+5)/(9-5) = - 3/4

Slope of tangent at B = - 1/m = - 1 / ( - 3/4) = 4/3

Answer 5

The slope of the radius CB is (-8-(-5))/(9-5)= -3/4
Thus the slope of the tangent is 4/3.

Answer 6

Center of circle (5,-5)

Line from Center to tangent point has slope ((-8)-(-5)) / (9 - 5) = -3/4

This line is perpendicular to the tangent.
Slope of twol lines that are perpendicular are negative reciprocals.
Slope of tangent line is -1/(-3/4) or 4/3

Answer 7

This is using analitical geometry
Since all tangen line on a circle always ortogonal againts the radius, so since the center is on (5,-5) and the point where the tangens line located is on B (9,-8), by using { (y-y1)/(y2-y1)} = { (x-x1)/(x2-x1)} // formula for finding a line equation.

We are looking for line that connecting Point C and B And the line equation is y = -3/4 x -5

and 2 ortogonal lines have characteristic that m1 * m2 = -1
// m1 = gradient line1 , m2 gradient line2

and we find that the m2 < the gradient thet we are looking for>
m2 = -1/m1
m2 = 4/3

// you can also using euclidian geometry to finish this and it's more easier but it's hard to expalin here because you need to see the figure.

Answer 8

Sines and cosines. Sort of like vectors (the tangent 'velocities' change back and forth with sines / cosines).

Answer 9

5!!! im no help!!