To their surprise the carriage was empty when they opened it, save a note left on the seat that said

]]>Now that you've understood the many concepts behind this problem (i.e. intersection of line and curve, gradients of perpendicular lines etc.), do note the circumstances in which they were used and try to look out for opportunities to apply them again in your future geometrical questions, where appropriate.

Meanwhile,

新年快乐!

Dawn brings forth another day ... and a faint whisper was heard again from that carriage ...

Yes, we can also get the same answer using Coordinate Geometry, which most of us would've learnt in Sec 3.

From the above diagram, and from your basic Geometrical Properties of Circles, we see that the radius is perpendicular to the tangent line `x` + `y` = 10.

So if we can obtain the coordinate of the intersection point between the tangent and the radius, we can then calculate the length of the radius (i.e. between the intersection point (`x`_{1}, `y`_{1}) and the circle's center (0, 6)) via the *Length of Line Segment* formula from your E-Maths Coordinate Geometry:

(1)

This is the equation of the tangent

A line passing through (0, 6) will intersect the line at right angles; perpendicular tangent to a circle. Yup yup - tangent ⊥ radius AMEN

hence

a line of gradient 1 passing through (0, 6) is:

Yes since:

- The radius ⊥ tangent ⇒ gradient of radius = 1 (∵
`m`_{1}`m`_{2}=-1) - The
`y`-intercept = 6 since the circle is centered at (0, 6)

∴ equation of the radius:

(2)

finding the point of intersection between and

Equating (1) and (2) to obtain (`x`_{1}, `y`_{1})

The point of intersection is ( 2, 8 )

find r, the distance from ( 0, 6 ) to ( 2, 8 )

Yes, and you don't even need to know the equation of the circle to get the answer!

I was thinking of using implicit differentiation at first too... Haha.

Don't stress our poor Sec Three rescuer (with A-Level stuff) already *lah* 😛 Miss Loi hasn't eaten in two days!

The ~~KBKP~~ voice from the loud hailer brought an abrupt end to the silence.

]]>THANK YOU SO MUCH

SOMEONEFOR YOUR HELP!BUT I FAILED MY AMATHS AND DROPPED OUT OF SCHOOL BEFORE I REACHED SEC 4! I'VE NOT DONE EQUATION OF CIRCLES BEFORE SO I DON'T FOLLOW YOUR WORKINGS!

AS MY CHER ONCE TOLD ME, CANNOT ANYHOW TRUST ANSWERS WITHOUT

UNDERSTANDINGTHE WORKINGS! SO DO YOU KNOW ANOTHER APPROACH THAT A SEC 3 DROPOUT LIKE ME WILL UNDERSTAND?!

Miss Loi: ~~Oh please don't be angry at Miss Loi as she is in the process of marking and commenting on your workings so expect some amendment/re-amendment here and there in the meantime.~~

~~Is it ok with you if Miss Loi uses (1), (2) etc. to label the equations in your workings - instead of the ( ) which caused her some confusion in her carriage last night?~~

Okie done.

the length should be

please help me to correct my original post. =) ]]>

*Dawn has broken ... a weak voice could be heard coming from that dilapidated carriage swinging gently to the breeze ...*

**Miss Loi:** In a geometrical problem like this when the person who set it was too lazy (or just being sadistic) to include a diagram, it's always a good practice to draw a quick sketch in order to get your orientation right and know *exactly* what you're looking for (especially if you're one of those 'visual' type of students):

From the diagram above, it's pretty clear that the total rope length required = 6+`r` units, since Miss Loi is right at the top of the wheel. So the only unknown is `r` the radius of the wheel.

P.S. Just a quick sketch will do - please don't waste time drawing Miss Loi's face during your exam!

Miss Loi is really, really glad to see your approach using the following properties from the little *Intersection of Line & Curve Leading to a Quadratic Equation* section (though it wasn't what Miss Loi had in mind when she set this question - see comment #5 ...) :

Our scenario is a classic problem (with one unknown, `r` in our case) for this oft-missed portion that's tucked inside your Quadratic Equations chapter, as some students have the impression that discriminants are only used in cases regarding intersections on the `x`-axis.

(1)

the equation of the circle is:

(2)

When solving the 2 simultaneous equations, it's more convenient to sub `y`=10-`x` into the circle equation (2) as you've done ( vs equating (1)=(2) ), as the circle equation contains a `y`^{2}.

Sub `y`=10-`x` into (2):

since x+y=10 was a tangent, the

the ferris wheel is a bit small... but I graphed the equations and it seems valid.

Miss Loi didn't specify the scale of the axes, so each unit could be in meter, feet, km etc 😛 And you used GC!

~~length of rope is 2r, the diameter of the wheel to reach the top of the wheel,~~

**units**

Correction: should be `r`+6 (see comment #2 below) - that's why sketching a quick diagram will likely to help one to visualize the problem better 🙂