Project maths question

extremebogman

Can't get my little head around this one. From 2011 SEC Sample:

The centre of a circle lies on the line x+2y-6=0. The x-axis and y-axis are tangents to the circle. There are two cirlces that satisfy these conditions. Find their equations.

ChemHickey

extremebogman wrote: »

Can't get my little head around this one. From 2011 SEC Sample:

The centre of a circle lies on the line x+2y-6=0. The x-axis and y-axis are tangents to the circle. There are two cirlces that satisfy these conditions. Find their equations.

Draw a little diagram.

I'm on my phone so apologies in advance for typos.

if x axis is a tangent then g^2=c
if y axis is a tangent, then f^2 =c

if centre lies on the line therefore (-g,-f) satisfy the eqn of the line.

So you have a good few things to equate

g^2=f^2

Therefore g=f
or.
g=-f

you can find either by subbing into line you found eg. where -g, -f satisfy x + 2y -6

this means
-g -2f-6= 0


Equate and hopefully it'll work to dFind values for g f c.

also, r is equal to half the diameter which you might be able to read of your little diagram. but I'm not certain as I'm on my phone and didn't write it out!

apologies if there are mistakes.

lostatsea

http://www.studentxpress.ie/ProjectMaths/PMSamplePaper2Soln(2011)Q4.pdf

reznov

lostatsea wrote: »

http://www.studentxpress.ie/ProjectMaths/PMSamplePaper2Soln(2011)Q4.pdf

Don't forget about the radius formula either AND the distance formula. Frequently used to solve problems.

NumberZeroNine

This method is very easy to grasp once explained but as always you should be careful to show enough work so that it can be followed step by step by an examiner.

I will explain it in full and then offer how much you should write in the exam in a quote bubble at the bottom.

The maths required is JC ordinary level and it helps if you can visualize the angles and distances so a rough 20second sketch will stop you making silly mistakes.

...

Consider a circle with the y-axis as a tangent.
Let S = (x1,y1) be the center.
There is some point T (the point of tangency) on the circle where T=(0,y2).

Now because every tangent has to be perpendicular to the radius at it's point of tangency, and because we know this tangent is vertical, we know that the line between T and the center S (the radius to T) is horizontal.

This tells us that the y coordinates will be the same, or that their difference is zero.
y1=y2
This means the distance between T and S (the radius) is now a straightforward difference between their x coordinates.

dist(A,B)= Sqrt( sq(xb-xa) + sq(yb-ya) )
dist(T,S)= Sqrt( sq(x1-x2) + sq(y1-y2) )
dist(T,S)= Sqrt( sq(x1-x2) + (o) )
dist(T,S)= Sqrt( sq(x1-0) + (o) )
dist(T,S)= Sqrt( sq(x1) )
dist(T,S)= |x1| = r

So in short...

If the circle C has the y-axis as a tangent, the radius of C is equal to the absolute value of the x-coordinate of the center S

By an almost identical construction, we can show that the same is true for the y-coordinate of S if the x-axis is tangental to C.

So...

If the circle C has the x-axis as a tangent, the radius of C is equal to the absolute value of the y-coordinate of the center S

And if both the x-axis and y-axis are tangents, it follows directly from this that |g| = |f| = r [where S = (-g,-f)]

So g = f OR g = -f

And these will give us enough to find the two circles in this question:

S1 = (-g,-f) S1 e x+2y-6=0
S2 = (-g,f) S2 e x+2y-6=0

SAMPLE ANSWER

Circle C: x^2+y^2+2gx+2fy+c = 0
Center: S=(-g,-f)
Radius r= sqrt(g^2 + f^2 - c)
x=0 is tangental to C
y=0 is tangental to C
S ∈ x+2y-6=0

x=0 is tangental to C ∴ r = |f|
y=0 is tangental to C ∴ r = |g|
∴ |f| = g

S= (-g,-g) OR S = (-g,g)

i) S= (-g,-g)
(-g,-g) ∈ x+2y-6=0
(-g) +2(-g) -6 =0
-3g -6 = 0
g = -2

∴S = (2,2)

r=|g|
r= sqrt(g^2 + f^2 - c)

(2) = sqrt( (-2)^2+(-2)^2 - c)
4 = 4 + 4 - c
c = 4

C: x^2+y^2+2gx+2fy+c = 0
C: x^2+y^2-4x-4y+4=0


ii) S= (-g,g)
(-g,g) ∈ x+2y-6=0
(-g) +2(g) -6 =0
g -6 = 0
g = 6

∴S = (-6,6)

r=|g|
r= sqrt(g^2 + f^2 - c)

(6) = sqrt( (-6)^2+(6)^2 - c)
6= 6 + 6 - c
c = 6

C: x^2+y^2+2gx+2fy+c = 0
C: x^2+y^2-6x#+6y+6=0

All in all this would easily save even those going for 600 a precious 60sec in the exam just in time spent on procedural writing and arithmetic.

MathsManiac

Here's an even shorter way. It's basically the same, but doesn't bother with all that g,f,c stuff.

As other posters have said, both the x-coordinate and y-coordinate of the centre have to be (+/-) the radius. If you draw a quick sketch of the line x+2y-6=0, it becomes immediately obvious that the only possible locations for the centre are in the 1st and second quadrant, so the centre is either (r,r) or (-r,r). Substitute (r,r) into the line to get r=2 and sub (-r,r) in to get r=6.

So the two possible equations are (x-2)^2 + (y-2)^2 = 4 and (x+6)^2 + (y-6)^2 = 36.

(By the way, NumberZeroNine, I think your last line in the box has a couple of typos.)