I'm having a huge amount of trouble with Exercise 3A, questions 10, 11 and 12 from the Heinemann P3 book. As I've got the C2 exam on Monday, any help would be really appreciated. The questions are:
10) Show that the line with equation 2x - 3y + 26 = 0 is a tangent to the circle with equation x² + y² - 4x + 6y - 104 = 0. If T is the point of contact of the tangent with the circle, find the equation of the normal to the circle at T. Find also the coordinates of the point on the circle which is diametrically opposite to T. Answer: 3x + 2y=0; (8, - 12)
11) The line with equation y=mx is a tangent to the circle with equation x² + y² - 6x - 6y + 17 = 0. Find the possible values for m. Answer: (9±√17)/8
12) Prove that the circle with equation x² + y² - 2ax - 2ay + b² = 0 touches the y-axis.
Hence, or otherwise, find equations of the two circles which pass through the points (1,2) and (2,3) and which touch the y-axis. Find the distance between their centers. Answer: x² + y² - 10x + 2y + 1 = 0, x² + y² - 2x - 6y + 9 = 0; 4√2
Thank you so much for taking the time to help me!
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P3 Coordinate Geometry- The Circle watch
- Thread Starter
- 18-06-2005 12:17
- 18-06-2005 12:22
2x - 3y + 26 = 0
2x = 3y - 26
x = 3y/2 - 13
x² + y² - 4x + 6y - 104 = 0
(3y/2 - 13)² + y² - 4(3y/2 - 13) + 6y - 104 = 0
9y²/4 - 39y + 169 + y² - 6y + 52 + 6y - 104 = 0
13y²/4 - 39y + 117 = 0
b^2 - 4ac = 1521 - 1521 = 0
b^2 - 4ac = 0 indicates two equal roots, and that the line touches.
Basically you have to treat the two equations, the line and circle, as simultaneous equations, and show that the discriminant of the resulting quadratic in x or y (depending on which you chose to substitute) is equal to zero for touching once, greater than 0 for going through the circle, and less than 0 for missing the circle.
Ill get on the other ones..
- 18-06-2005 12:28
Oh, to finish that question off, solve the quadratic I finished with above to get the y value, then from that use the line equations to find the x value. Then you can use the gradient of the normal with this point to find the equation for the normal.
After that, you can use simultaneous equations with this new line and the circle to find the two points of crossover. One will be the one you already used, the second will be the point opposite.
- 18-06-2005 12:39
11) The line with equation y=mx is a tangent to the circle with equation x² + y² - 6x - 6y + 17 = 0. Find the possible values for m. Answer: (9±√17)/8[/QUOTE]
y = (mx)
x² + y² - 6x - 6y + 17 = 0
x² + (mx)² - 6x - 6(mx) + 17 = 0
x² + m²x² - 6x - 6mx + 17 = 0
x²(1 + m²) - x(6 + 6m) + 17 = 0
b^2 - 4ac = 0 for tangent to the circle,
(6 + 6m)² - 4(1 + m²)(17) = 0
36 + 72m + 36m² - 68 - 68m² = 0
-32m² + 72m -32 = 0
4m² - 9m + 4 = 0
m² - 9m/4 = -1
(m - 9/8)² = 81/64 - 1
m - 9/8 = ±√(17/8)
m = (±√17 + 9)/64
- Thread Starter
- 18-06-2005 12:48
Thank you so much! It all makes perfect sense and I'm just wondering why I didn't think of it before; especially the first one- I thought that you needed to differentiate the equation and so on.
I doubt that I can help you in return, but if you do think of anything, just shout.