Locus and Subsets (1 Viewer)

[kbei]

Hey guys, just doing a little extra here and im pretty stuck on this:

11 (a) Find the equation of a circle whose centre is the point (-1,2) and whose radius has a length of 5 units. I could work this question out which turned to be:
x^2+y^2+2x-4y-20=0

But part b had me stumped:
b) What is the length of the intercept cut off by this circle on the X-axis?
I dont understand what this question is asking, is it asking for the distance between the 2 points in which this circle touches the X-asix? if so, how do i find the 2 points, if not then ignore this.

c) FInd the length of the tangent from the point (6,4) to this circle.
I'm not sure which point this question is referring to....

Thanks for your help and time

 

[Trebla]

Hey guys, just doing a little extra here and im pretty stuck on this:

11 (a) Find the equation of a circle whose centre is the point (-1,2) and whose radius has a length of 5 units. I could work this question out which turned to be:
x^2+y^2+2x-4y-20=0

But part b had me stumped:
b) What is the length of the intercept cut off by this circle on the X-axis?
I dont understand what this question is asking, is it asking for the distance between the 2 points in which this circle touches the X-asix? if so, how do i find the 2 points, if not then ignore this.

c) FInd the length of the tangent from the point (6,4) to this circle.
I'm not sure which point this question is referring to....

Thanks for your help and time

The equation of the circle is (x + 1)² + (y - 2)² = 25

b) I think the question wants the distance between the two x intercepts. In which case sub y = 0
(x + 1)² = 21
x = - 1 ± √21
So the distance between them is: - 1 + √21 - (- 1 - √21) = 2√21 units

c) The question requires the length of a tangent from an external point.
The quickest way to do it is using a circle geometry property which is not part of the 2 unit course...
When you a draw a tangent to circle, it is perpendicular to the radius of that circle. This is a theorem from Extension 1. Without that I think you would have to find the point of contact which is annoying...lol

So using that theorem, you can form a right-angled triangle with the point of contact, and the points (6,4) and (-1, 2). Let the distance from (6,4) to the point of contact be d.

By Pythagoras' theorem: 5² + d² = [√(6 + 1)² + (4 - 2)²]²
25 + d² = 53
d² = 28
.: d = 2√7 (d > 0)

 

[kbei]

thanks trebla for that, now i get it but i got stuck on another question:

"A" is a point where the circle with equation x^2+y^2=16 cuts the X-axis. Find the locus of the midpoints of all chords of this circle that contain the point A.
I'm not sure where to start, and a diagram would be really helpful as I am a visual learner. I tried to draw it up, but it sort of became stuffed....thanks

 

[Timothy.Siu]

i think i got the answer, but i used sorta 3unit stuff
i'm sure theres a 2unit way if this is a 2unit question

i let A be (4,0)
and some other point be Q(s,+-sqrt(16-s^2)) -4<s<4

midpoint of QA is ((4+s)/2,+-sqrt(16-s^2)/2)

x=(4+s)/2
y=+-sqrt(16-s^2)/2

s=2x-4
sub it into y, and simplifying...
y=+-sqrt(4x-x^2)

 

[clintmyster]

havent really solved this yet but if i was to take a guess id say the locus is two circles one of centre 2,0, the other -2,0 and they each have a radius of 2 units

 

[Timothy.Siu]

havent really solved this yet but if i was to take a guess id say the locus is two circles one of centre 2,0, the other -2,0 and they each have a radius of 2 units

yeah...lol, u can tell if u draw a diagram...but i sorta solved it

 

[kbei]

yeah. er..not sure what the diagram even looks like, just started this topic, and as you can see, its not my strongest point, really appreciate if someone could give a pointer on this one

 

[Timothy.Siu]

heres a diagram, ur trying to find the locus of the orange dots.

 

[kbei]

Thanks Timothy.Siu, i worked out the question nearly instantly after i saw that circle, but i hit another brick wall with this topic, the question is:

Find the equation of the locus of the midpoints of all chords of length 4 of the curcle with equation x^2+y^2-4x+2y=4. I worked this question out by drawing it out, but i was told that is not the way to solve these questions. Could someone provide the theory side of the question with working out? thanks in advance

 

[Trebla]

Thanks Timothy.Siu, i worked out the question nearly instantly after i saw that circle, but i hit another brick wall with this topic, the question is:

Find the equation of the locus of the midpoints of all chords of length 4 of the curcle with equation x^2+y^2-4x+2y=4. I worked this question out by drawing it out, but i was told that is not the way to solve these questions. Could someone provide the theory side of the question with working out? thanks in advance

Diagrammatic methods usually simplify the problem and make it easier to visualise so you know what type of answer to expect. The algebraic methods are for finding the equations. There is no one rule you have to follow.
Diagrammatically you should expect the locus to be a circle.
x² + y² - 4x + 2y = 4
=> (x - 2)² + (y + 1)² = 9
This is a circle of radius 3 and centre (2, -1)
A chord of length 4 will form an isosceles triangle with the radius.
So a perpendicular bisector of this chord will form two congruent right angle triangles of sides 3, 2 and √5.
So no matter where the point is, the distance from the centre to the chord is √5.
If (x,y) is a variable midpoint of the chord, then the locus is a circle of the same centre but with radius √5
i.e. (x - 2)² + (y + 1)² = 5
(this can be derived from the distance formula from the centre to the variable point)