# P3 CirclesWatch

This discussion is closed.
#1
P3 Heinemann Review Exercise 1 Q31
Find the coordinates of the center and the radius of the circle with equation x² + y² -10y = 0.
Prove that the line with equation 4x - 3y + 40 = 0 is a tangent to the circle.

For the first part I got center (0, 5) and radius 5, which is correct. I can't seem to do the second part.

I tried to prove that the line intersects the circle at only 1 point by substituting y=(4x+40)/3 into the equation of the circle, but that didn't work. I then tried substituting x=(3y-40)/4, in case I'd made an arithmetic slip in the first subsitution, but that didn't work either.

Help?
0
14 years ago
#2
(Original post by shift3)
P3 Heinemann Review Exercise 1 Q31
Find the coordinates of the center and the radius of the circle with equation x² + y² -10y = 0.
Prove that the line with equation 4x - 3y + 40 = 0 is a tangent to the circle.
.
?
x=(3y-40)/4
x^2=9y^2-240y+1600/16
so 9y^2-240y+1600+16y^2-160y=0
25y^2-400y+1600=0
y^2-16y+64=0
(y-8)^2=0
so y=8
x=-4
0
14 years ago
#3
(Original post by shift3)
P3 Heinemann Review Exercise 1 Q31
Find the coordinates of the center and the radius of the circle with equation x² + y² -10y = 0.
Prove that the line with equation 4x - 3y + 40 = 0 is a tangent to the circle.

For the first part I got center (0, 5) and radius 5, which is correct. I can't seem to do the second part.

I tried to prove that the line intersects the circle at only 1 point by substituting y=(4x+40)/3 into the equation of the circle, but that didn't work. I then tried substituting x=(3y-40)/4, in case I'd made an arithmetic slip in the first subsitution, but that didn't work either.

Help?
make them equal to each other...
0
#4
omg.. I wrote down x²-y²-10y=0 as the equation of the circle on the pad I was working on. No wonder I couldn't prove it!

Thanks mathz!
0
14 years ago
#5
(Original post by shift3)
omg.. I wrote down x²-y²-10y=0 as the equation of the circle on the pad I was working on. No wonder I couldn't prove it!

Thanks mathz!

Get the intersection point (see mathz above)
the gradient of the line is 4/3
you can prove that the gradient of the radius between (0,5) and (-4,8) is
-3/4, so the radius and the line are at rt angles to each other, so the line is a tangent

Aitch
0
#6
(Original post by Aitch)
Get the intersection point (see mathz above)
the gradient of the line is 4/3
you can prove that the gradient of the radius between (0,5) and (-4,8) is
-3/4, so the radius and the line are at rt angles to each other, so the line is a tangent

Aitch
That's not really required. If I manage to show that there's only one point of intersection between the line and the circle, then it's a tangent.
0
14 years ago
#7
yp teh 2nd post is cul - but got to add, that thi sis only true is there is only one real and equal root, can't have imaginary or two roots to the final equation - if there is only 1 real and equal root, then its self evident that its gonna be a tanget, cos this is one of its definitions!!!
0
14 years ago
#8
(Original post by shift3)
That's not really required. If I manage to show that there's only one point of intersection between the line and the circle, then it's a tangent.
You are, of course, correct!

I've forgotten more of my P3 than I thought! What I did was establish the gradient of a tangent of which we already had the equation...

...Good procedure,wrong question!

Aitch
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