Need Help with Locus Qns (1 Viewer)

[ameh]

I'm having a little difficulty with these locus questions [from grove], any help would be appreciated. btw I have the answers, just not sure about the working out

17. Prove that the line 3x+4y+21=0 is a tangent to the circle [x^2]-8x+[y^2]+4y-5=0

22. Find the equation of the locus of a point that moves so that it is equidistant from the line 4x-3y+2=0
and the line 3x+4y-7=0

23. Find the equation of the locus of a point that moves so that it is equidistant from the line 3x+4y-5=0
and the line 5x+12y-1=0

24. A circle has centre C(-1,3) and radius 5units:
(b) The line 3x-y+1=0 meets the circle at two points. Find their co-ordinates

25. (a) Find the perpendicular distance from P(2,-5) to the line 5x+12y-2=0
(b) hence find the equation of the circle with centre P and the tangent 5x+12y-2=0


Thanks =)

 

[Trev]

17. Prove that the line 3x+4y+21=0 is a tangent to the circle [x^2]-8x+[y^2]+4y-5=0

Factorise the circle so you can find it's centre and radius.
Use that formula that finds out perpendicular distance from the centre of the circle to the line 3x+4y+21=0. It will be a tangent if both the radius and the perp. distance are equal.

 

[Trev]

22. Find the equation of the locus of a point that moves so that it is equidistant from the line 4x-3y+2=0 and the line 3x+4y-7=0

I'm so shit I can't remember how to do this, I think. You can't blame me, last time I did something like this was halfway through year 11!
I <u>think</u> you use the perpendicular formula using one line and a point, say P(x₁,y₁), and make that equal to the other line using the perp. formula.....hope so haha. If this works out, just do the same for the other question too, they're both the same method.

 

[rama_v]

17. You do this one by first rearranging the equation of the circle:

x^2 - 8x +y^2 +4y -5=0
complete the square for both x^2 and y^2
[x-4]^2 -16 + [y+2]^2 -5=5
[x-4]^2 + [y+2]^2=25

so you have a circle with centre (4, -2) and radius 5. To prove its tangent use the perpendicular distance formula, the point is (4, -2) and the equation of the line is 3x+4y+21=0

D=[3(4)+4(-2) +21]/sqrt(3^2 + 4^2) = 5

therefore its a tangent to the circle since the distance from teh centre is the same as the circle's radius.

 

[Trev]

24. A circle has centre C(-1,3) and radius 5units:
(b) The line 3x-y+1=0 meets the circle at two points. Find their co-ordinates

Equation of circle is (x+1)² + (y-3)² = 25; and just sub in the equation of the line into that.
So y=3x+1.
(x+1)² + (3x-2)² = 25, expand that and find you x coordinates of the intersection, and hence ones for y.

 

[Trev]

25. (a) Find the perpendicular distance from P(2,-5) to the line 5x+12y-2=0
(b) hence find the equation of the circle with centre P and the tangent 5x+12y-2=0

Use the perp. distance formula with the point (2,-5) and the line given, you should be able to do this, it's only substitution.
since this will be the radius of the circle, just sub that into the equ. of the circle: (already putting in centre point wateva):
(x-2)² + (y+5)² = r²

 

[Dreamerish*~]

23. Find the equation of the locus of a point that moves so that it is equidistant from the line 3x+4y-5=0
and the line 5x+12y-1=0

use perpendicualr distance formula:

| ax₁ + by₁ + c |​

./ a² + b²​

3x + 4y - 5 = 0 ...... 1
5x + 12y - 1 = 0 ...... 2

let P be the points x and y which make up the locus.
i.e. P(x, y)

dP1 =

| 3x + 4y - 5 |​

./ 3² + 4²​

=

3x + 4y - 5 ​

5​

(i'm not 100% sure about this bit)

similarly, dP2 =

| 5x + 12y - 1 |​

./ 5² + 12²​

=

5x + 12y - 1 ​

13​

because dP2 = dP1
therefore (5x + 12y - 1)/13 = (3x + 4y - 5)/5
13(3x + 4y - 5) = 5(5x + 12y - 1)
39x + 52y - 65 = 25x + 60y - 5

14x - 8y - 60 = 0 is the equation of the locus.

 

[Trev]

Dreamerish*~; after putting the equations of both lines to begin with, and the equation of locus your worked out into Graphmatica, it really doesnt look right! lol....

 

[KFunk]

Putting her results into graphmatica it does look right .

If you take a negative value for one of the perp. distance formulas ('cause of the absolute value) you get another line:

- 39x - 52y + 65 = 25x + 60y - 5
64x +112y -70 = 0

You can see visually with graphmatica, as with the other line, that it satisfies the conditions of the locus.

Does anyone know if the 2U locus allows for this answer or does it simply want the first one?

 

[Dreamerish*~]

yeah that's what i was thinking... i wasn't sure if i should have taken both values or not. i guess i should have since there are absolute value signs.

what's graphmatica? is it like winplot?

 

[KFunk]

Yeah, same deal. Cheap, free graphing program.

 

[ameh]

thanks guys!

i might just go download Graphmatica...

 

[ameh]

ok I have more questions from Fitz:

8. Twice the distance of a point from the x-axis is 3 times its distance from the y-axis

9. The distance of a point from the line y=-5 is three quarters of its distance from the line x=2

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

8. (c) For circle x^2 + y^2 + 6x-8y=0 Find the diameter drawn through the origin.

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

7. Show equation x^2 + y^2 -6x+2y+10=0 represents a point circle. Btw what is a point-circle?


=)

 

[Trev]

Putting her results into graphmatica it does look right .

Haha i'm so shit. End story.

 

[Trev]

8. Twice the distance of a point from the x-axis is 3 times its distance from the y-axis

Both points would be (x,0) and (0,y)
so its:
2./[(x-x)² + (y-0)²] = 3./[(x-0)² + (y-y)²]
4x²=9y²

PS. './' means square root

 

[m_isk]

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

feelin lazy so...for this one, just sub in x=4 and y=-3 into the equation and show that it DOES NOT equal to zero...

 

[Dreamerish*~]

7. Show equation x^2 + y^2 -6x+2y+10=0 represents a point circle. Btw what is a point-circle?

x² + y² - 6x + 2y + 10 = 0

x² - 6x + 9 - 9 + y² + 2y + 1 - 1 + 10 = 0

(x - 3)² - 9 + (y + 2)² - 1 + 10 = 0

(x - 3)² + (y + 2)² = 0

the formula for a circle is (x + h)² + (y + k)² = r² where (h, k) is the centre, and r is the radius.

in (x - 3)² + (y + 2)² = 0, r = 0 and therefore the radius is 0. hence the circle is called a point-circle. it's just a dot

i didn't know this either until i worked it out.

 

[Trev]

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.

 

[Trev]

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

 

[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