Suppose you know two facts:
1) the equation of a circle: x^2 + y^2 = 64
2) the coordinates of a single point on a line tangent to the circle: (12,0)
How do you find the slope of the tangent line?
Suppose you know two facts:
1) the equation of a circle: x^2 + y^2 = 64
2) the coordinates of a single point on a line tangent to the circle: (12,0)
How do you find the slope of the tangent line?
Easy enough. So it lies on a circle. So you have a radial line going to that point on the circle (12,0). It's slope is just the y coordinate over the x-coordinate. The tangent line will be perpendicular to this line.
Use implicit differentiation.
d/dx the equation
you get
2x+2ydy/dx=0
solve for dy/dx
dy/dx= -x/y
plug in values for for x and y, in this case y=0 so you have a vertical line
another d bag wrote:
d/dx the equation
you get
2x+2ydy/dx=0
solve for dy/dx
dy/dx= -x/y
This much is right, but (12, 0) is not on the circle.
zip zero nada wrote:
Easy enough. So it lies on a circle. So you have a radial line going to that point on the circle (12,0). It's slope is just the y coordinate over the x-coordinate. The tangent line will be perpendicular to this line.
The point (12, 0) is NOT on the circle. It's on the line tangent to the circle.
P.S. wrote:
another d bag wrote:d/dx the equation
you get
2x+2ydy/dx=0
solve for dy/dx
dy/dx= -x/y
This much is right, but (12, 0) is not on the circle.
don't know what d/dx is... is d for delta?
Since dy/dx = -x/y for a point on the circle, I set that equal to the slope of a line containing a point (x, y) on the circle and the point (12, 0), giving me -x/y = (12 - y)/(-x).
Multiply both sides by -xy and you've got x^2 = 12y - y^2, or x^2 + y^2 = 12y. Along with the original circle, this means the y coordinate of the point on the tangent line that touches the circle is plus or minus 16/3, and it follows that the x coordinate that goes with them both is 320/9. Put that point and the point (12, 0) into your good old slope formula and that's the slope of your tangent line.
d/dx is notation for taking the derivative with respect to x.
Did you get it?
why am i so dumb? wrote:
Suppose you know two facts:
1) the equation of a circle: x^2 + y^2 = 64
2) the coordinates of a single point on a line tangent to the circle: (12,0)
How do you find the slope of the tangent line?
The point of tangency (P) is at a distance of 8 from (0,0).
The points (0,0), P, (12,0) form a right triangle with sides of length 12, 8 and L.
Since a^2 + b^2 = c^2 we get L^2 + 64 = 144. L = SQRT(80) = 4*SQRT(5).
Call the point P (X,Y). Then the points (0,0), (X,0), and P form a right triangle that is similar to the triangle formed by (0,0), P, (12,0). As a result Y/X = L/8 = SQRT(5)/2.
Note that Y/X can also be -SQRT(5)/2 (the point of tangency lies below the x-axis)
Now go do the rest of your homework.
I'm sorry, that should have been sqrt(320)/3 instead of 320/9.
A right triangle is formed with a radius of the circle, it's tangent, and the hypotenuse is 12.
(8,80^.5,12) triangle
Find the vertical bisector of that triangle.
The newly formed outer triangle is ((320/9)^.5,20/3,80^.5)
Slope of tangent line is rise/run of above^: ~5.96/6.66 = .89 (there are two tangent lines, so -.89 as well)
Nobody wants to do your HW 4 U wrote:
The points (0,0), P, (12,0) form a right triangle with sides of length 12, 8 and L.
How do you figure?
none of your answers are right
if you know the answer, why are you asking?
why am i so dumb? wrote:
Suppose you know two facts:
1) the equation of a circle: x^2 + y^2 = 64
2) the coordinates of a single point on a line tangent to the circle: (12,0)
How do you find the slope of the tangent line?
Some advice:
In general, the best way to approach a problem like this is to get a large blank piece of paper and a pencil, and draw the picture in some realistic way. Then start drawing lines and triangles until you understand why the question makes sense. From there, you can start labeling (X,Y) coordinates and use your knowledge of geometry to change the geometry problem into a mathematics problem. Once you have the right equations written down (this is the hard part!), the math itself will be easy to solve.
If you do this, I think you'll see that your question as more than one answer (I'm picturing the problem in my head and it seems that way to me), I think there should be two tangent lines.
why are u so smart wrote:
if you know the answer, why are you asking?
maybe he has been told the correct answer and needs to figure out how to get it.
it's also possible that he gave us the wrong problem.
why are u so smart wrote:
if you know the answer, why are you asking?
the answer is in the back of the book
P.S. wrote:
Since dy/dx = -x/y for a point on the circle, I set that equal to the slope of a line containing a point (x, y) on the circle and the point (12, 0), giving me -x/y = (12 - y)/(-x).
shouldn't the slope of the tangent line be (12 - x)/(-y) instead of (12 - y) / (-x)? Can you explain?