wh wrote:
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?
3 non-linear pts form a triangle. It is a right triangle because the line defined by P and (12,0) is a tangent line to the circle while the line defined by (0,0) and P is a radius. Lengths of 12 and 8 are given.
Nobody wants to do your HW 4 U wrote:
3 non-linear pts form a triangle. It is a right triangle because the line defined by P and (12,0) is a tangent line to the circle while the line defined by (0,0) and P is a radius. Lengths of 12 and 8 are given.
It seems to me that it could only be a right triangle if P was on the y axis such that the line from P to the origin made a right angle with the line fromt he origin to (12, 0).
why am i so dumb wrote:
Nobody wants to do your HW 4 U wrote:3 non-linear pts form a triangle. It is a right triangle because the line defined by P and (12,0) is a tangent line to the circle while the line defined by (0,0) and P is a radius. Lengths of 12 and 8 are given.
It seems to me that it could only be a right triangle if P was on the y axis such that the line from P to the origin made a right angle with the line fromt he origin to (12, 0).
OK, at this point I am wondering if you are kidding.
But assuming not...
Any line of radius is perpendicular to a line that is tangent to the circle at the intersection of the radius with the circle.
Thus the line {(0,0), P} is perpendicular to the line {P, (12,0)}
First draw the tangent line from (12,0) to the circle. (There's actually two of them, though the magnitude for them is the same, but one of the tangent lines has positive slope, and the other has negative). Next, draw the radius 8 to the point where the tangent line is tangent to the circle. This is now a right triangle with leg 8 and hypotenuse 12, so by the Pythagorean Theorem, the other leg is sqrt(80). This is the distance from (12,0) to the tangent point. Then, drop a line from the tangent point to the x-axis, call this "y". We then call the distance from the center of the circle (the origin) to the intersection of the dropped line and the x-axis "x". The distance from the point of intersection between the dropped line and the x-axis and (12,0) is "12-x". So we now have two right triangles. One with legs x and y and hypotenuse 8. The other with legs 12-x, y, and sqrt(80). So solve this system of equations:
x^2 + y^2 = 8^2
(12-x)^2 + y^2 = (sqrt(80))^2
(Solve for y^2 in the first one and then plug into the second) and solve for x. We get x = 5.33, and y = 5.96. This (x,y) coordinate is the tangent point. We now have two points on the tangent line. Slope = (5.96 - 0) / (5.33 - 12). So the answer is -1.2. (All of this assumes you initially drew your tangent line to a point in the 1st quadrant).
Could you possibly give us the answer? It's easier to backtrack in my opinion.
Math Prof. wrote:
First draw the tangent line from (12,0) to the circle. (There's actually two of them, though the magnitude for them is the same, but one of the tangent lines has positive slope, and the other has negative). Next, draw the radius 8 to the point where the tangent line is tangent to the circle. This is now a right triangle with leg 8 and hypotenuse 12, so by the Pythagorean Theorem, the other leg is sqrt(80). This is the distance from (12,0) to the tangent point. Then, drop a line from the tangent point to the x-axis, call this "y". We then call the distance from the center of the circle (the origin) to the intersection of the dropped line and the x-axis "x". The distance from the point of intersection between the dropped line and the x-axis and (12,0) is "12-x". So we now have two right triangles. One with legs x and y and hypotenuse 8. The other with legs 12-x, y, and sqrt(80). So solve this system of equations:
x^2 + y^2 = 8^2
(12-x)^2 + y^2 = (sqrt(80))^2
(Solve for y^2 in the first one and then plug into the second) and solve for x. We get x = 5.33, and y = 5.96. This (x,y) coordinate is the tangent point. We now have two points on the tangent line. Slope = (5.96 - 0) / (5.33 - 12). So the answer is -1.2. (All of this assumes you initially drew your tangent line to a point in the 1st quadrant).
The first correct solution.
Math Prof. wrote:
First draw the tangent line from (12,0) to the circle. (There's actually two of them, though the magnitude for them is the same, but one of the tangent lines has positive slope, and the other has negative). Next, draw the radius 8 to the point where the tangent line is tangent to the circle. This is now a right triangle with leg 8 and hypotenuse 12, so by the Pythagorean Theorem, the other leg is sqrt(80). This is the distance from (12,0) to the tangent point. Then, drop a line from the tangent point to the x-axis, call this "y". We then call the distance from the center of the circle (the origin) to the intersection of the dropped line and the x-axis "x". The distance from the point of intersection between the dropped line and the x-axis and (12,0) is "12-x". So we now have two right triangles. One with legs x and y and hypotenuse 8. The other with legs 12-x, y, and sqrt(80). So solve this system of equations:
x^2 + y^2 = 8^2
(12-x)^2 + y^2 = (sqrt(80))^2
(Solve for y^2 in the first one and then plug into the second) and solve for x. We get x = 5.33, and y = 5.96. This (x,y) coordinate is the tangent point. We now have two points on the tangent line. Slope = (5.96 - 0) / (5.33 - 12). So the answer is -1.2. (All of this assumes you initially drew your tangent line to a point in the 1st quadrant).
Here is a picture to illustrate the point I am stuck on:
http://www.aww-kittah-aww.com/up/public/191992/math.jpgLike I said, (16/3, sqrt(320)/3) and (16/3, -sqrt(320)/3) are the points of tangency. The slopes of the lines are -2/sqrt(5) and 2/sqrt(5).
why am i so dumb wrote:
Here is a picture to illustrate the point I am stuck on:
http://www.aww-kittah-aww.com/up/public/191992/math.jpg
Good that you drew it. It's a right angle at the point P -- the tangent line and the radial line must be perpendicular to each other, by definition.
why am i so dumb wrote:
Here is a picture to illustrate the point I am stuck on:
http://www.aww-kittah-aww.com/up/public/191992/math.jpg
Forgive me if this has been posted.
1) Use the graph you have.
2) The right triangle formed between the origin, (12,0) and the point on the circle you are seeking has a hypotenuse of length 12 and a leg of length 8 due to the circle. The line from (12,0) to the point in question has a length of 80^0.5.
3) Use the line with the length of 80^0.5 as the hypotenuse of a right triangle with one of the legs dropping from the point in question perpendicular to the x axis.
4) Solve the right triangle equation for the triangle in step 3. 80=(12-x)^2+y^2.
5) Substitute y^2=64-x^2 into the eqn. in step 4 since you know the point must lie on the circle and you can find y and then x.
6) Double check that the resulting (x,y) coordinate is correct. Find the angles that 2 lines form with the x axis, one line from the origin through the solved x,y point and one line from (12, 0) through the solved x,y point. If these 2 angles don't add to 90 degrees, try again.
From this diagram:
http://www.aww-kittah-aww.com/up/public/191992/math.j
Just use right triangle trig to get the angle made by the radius and the hypotenuse.
It would be arccos(8/12) = angle.
Then find the coordinates of P: (8cos(angle), 8sin(angle))
Remember the circle has radius 8.
Now you have two points on that tangent line.
Easy to get slope from there.
P.S. wrote:
Use implicit differentiation.
If you want to do 3x as much work, go ahead.
why am i so dumb wrote:shouldn't the slope of the tangent line be (12 - x)/(-y) instead of (12 - y) / (-x)? Can you explain?
yes
the OP gave the right method but unfortunately juxtaposed x & y ( easy to do )
following that method,
x = 5 1/3 , y = 320^0.5/3 or 5.9628
& slope is
+/- 0.89442
