Need Help with Locus Qns (1 Viewer)

[shafqat]

for 5, substitute the pt, lhs does not equal to 0, so it doesnt lie on circle

^ already said

 

[Dreamerish*~]

9. Find the equation of the circle that touches the x-axis at (4,0) and the Y-axis at (0,4)
You can see that the centre of the circle will be at (4,4) with a radius of 4. So;
(x-4)² + (y-4)² = 4²
expand etc

... uh oh.

 

[Trev]

8. (c) For circle x^2 + y^2 + 6x-8y=0 Find the diameter drawn through the origin.
simplify, so;
(x+3)² -9 + (y-4)² -16 = 0
(x+3)² + (y-4)² = 5²
Find the diameter drawn through the origin, do you mean distance or the equation?
Then use two point formula with centre circle and origin, so the points (-3,4) (0,0).

 

[shafqat]

for 8c, the centre is -3,4. so find equation of line going through this and origin

edit: damn, not again

 

[Dreamerish*~]

9. Find the equation of the circle that touches the x-axis at (4,0) and the Y-axis at (0,4)

this means that the radius of the circle is 4 units, and the centre is at the origin - (0, 0)
(i'm assuming this by picturing the circle in my head )

therefore with the equation (x + h)² + (y + k)² = r², in this case: h = 0, k = 0 and r = 4

hence the equation becomes: x² + y² = 16

i think this looks right....

 

[Trev]

... uh oh.

Haha don't worry, I thought there would be 2 possible equations for the circle at first, then realised one of them would produce more than just a tangent at those points.

 

[Dreamerish*~]

5. Show (4,-3) does not lie on circle x^2 + y^2 -5x+3y+2=0.

the equation is: x² + y² - 5x + 3y + 2 = 0

subbing in the values for x and y from (4, -3):

4² + (-3)² - 5(4) + 3(-3) + 2 = 0

16 + 9 - 20 - 9 + 2 = -2 and not 0

if the answer was 0, then the point (4, -3) would satisfy the equation and therefore lie on the circle.

it does not equal zero, hence it does not lie on the circle

 

[Trev]

Well I suppose Dreamerish*~ you are also right, since it doesn't specify they are the only points of intersection on either axis.... But (x-4)² + (y-4)² = 4² is definately a correct answer...

 

[Dreamerish*~]

Well I suppose Dreamerish*~ you are also right, since it doesn't specify they are the only points of intersection on either axis.... But (x-4)² + (y-4)² = 4² is definately a correct answer...

oh yeah now i see it
your one "touches" the axises (lol whatever they're called in plural)
mine crosses.

 

[ameh]

thanks guys!

 

[ameh]

9. The distance of a point from the line y=-5 is three quarters of its distance from the line x=2
(x,-5) and (2,y)
./[(x-x)² + (y+5)²] = ³/₄./[(x-2)² + (y-y)²]
16(y+5)²=3(x-2)² expand and etc.

Hey, I'm doing that but the answer at the back of Fitz is different. There's two equations !

 

[Trev]

Well, I don't know. No one has seemed to have corrected me, so maybe the answer in the back is wrong?!

 

[shafqat]

|y+5| = 3/4|x-2|
go from there i think

 

[KFunk]

9. The distance of a point from the line y=-5 is three quarters of its distance from the line x=2
(x,-5) and (2,y)
./[(x-x)² + (y+5)²] = ³/₄./[(x-2)² + (y-y)²]
16(y+5)²=3(x-2)² expand and etc.

I'm pretty sure that the distance of a point (x,y) from the line y=-5 is |y+5| and that the distance of (x,y) from x=2 is just |x-2|

so |y+5| = 3/4|x-2|

-----> 3x -4y -26 = 0 or 3x +4y +14=0 (EDIT: As stated above )

 

[Trev]

Hmmm, I have never (ever) used absolute values in questions such as this. But it sounds right, dangt.

 

[ameh]

argh i see why, i put 4 on the other side and squared it all

=/

sorry my bad!

 

[KFunk]

Hmmm, I have never (ever) used absolute values in questions such as this. But it sounds right, dangt.

You don't need to use absolute values, it just seems a quicker way to generate both solutions and I geuss it seems 'proper' when you're dealing with distances. Otherwise you could just use (x-2) and (2-x) as values for the distance between (x,y) and the line x=2 [same deal as using absolute value] and you would get the same two answers.