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Topic review (newest first)

bob bundy
2013-07-30 19:28:24

Above edited.

bob bundy
2013-07-30 16:23:27

hi cooljackiec,

If you collect together the x terms and add an amount you can make a perfect square.  Do the same for y and you'll end up with an equation like this:

a circle with centre at (a,b) and radius r the value of which you will know.

If you choose a point outside the circle you will have a value for x and y that cannot work in the equation. So think about the largest x valuie that will give a possible y.

bobbym: I'm getting a larger value than yours.

LATER EDIT:  Whilst sitting in the dentist's chair I re-worked this and realised my error.  I forgot to divide by 2 when completing the square.  I now get the same answer.  My apologises.  shame

Good news: teeth all OK.  smile


cooljackiec:  I have added a new post to an old question of yours at … 35#p279935

2013-07-30 05:44:17

Hi cooljackiec;

2013-07-30 04:47:06

Find the largest real number x for which there exists a real number y such that x^2 + y^2 = 2x + 2y.

bob bundy
2013-07-27 17:41:44

17/4 is what I got too.  smile

x by 4

shortcut note:  using the usual quadratic notation the midpoint is at -b/2a


2013-07-27 06:14:00

oh thanks. so it is the chord with (2,0) and (0,8) and the tangent. the radius is 17/4.

Last : You can substitute that value of y into the circle equation to solve for x.  As it's a quadratic you'll get two answers, one is A the other is B
I got the answer (0.4,0.8), but it is still wrong. i substituted, made an equation, and kept solving. i did the problem again, and i go the same result, but it is still wrong

bob bundy
2013-07-27 04:37:58

post 7 Q1


2013-07-27 02:06:04

aggh... i thought 25 + a = 36 and a was 9. the last coordinate is 11/5.

i got the bisecting square problem. the answer was -5/2.

for the 2nd one you tell me to use perpendicular bisectors. how?

bob bundy
2013-07-26 22:08:14

No.  You had 33/5 correct before.  It's the x coord that was wrong.

Let's check the steps.

What did you have for R ?


2013-07-26 08:34:45

is the x coordinate for the reflection problem 9/5??

bob bundy
2013-07-26 07:38:33

Are you thinking it must go through (0,0)  ?

It doesn't have to.

Find the centre of the square and substitute those coords into y = 6x + c


2013-07-26 06:20:15

for the square, how can a line with slope 6 bisect the square?

bob bundy
2013-07-26 04:03:14

cooljackiec wrote:

(7/5, 33/5) i believe.

It looks like you did the correct working and then wrote down the wrong answer.

33/5 is correct but 7/5 is not.  This is the x difference between P and R not the coord of Q.  My diagram is accurate.

Next question.  To bisect a square you would have to draw a line through its centre.  That should be enough of a hint.

Next : You can find the centre of the circle by using the perpendicular bisectors again.

Last : You can substitute that value of y into the circle equation to solve for x.  As it's a quadratic you'll get two answers, one is A the other is B


2013-07-26 02:03:24

(7/5, 33/5) i believe.

A line with slope 6 bisects the area of a unit square with vertices (0,0), (1,0), (1,1), and (0,1). What is the y-intercept of this line?

A circle is tangent to the y-axis at the point (0,2) and passes through the point (8,0), as shown. Find the radius of the circle.

The line y = (x - 2)/2 intersects the circle x^2 + y^2 = 8 at A and B. Find the midpoint of \overline{AB}. Express your answer in the form "(x,y)."

bob bundy
2013-07-25 03:18:12


Yes, I think 20/3 is correct, well done!

Here's a way to do the next one.  (see diagram)

Find the equation of PQ.  You know its slope is -2 and it goes through (5,1)

Then find where it crosses the other line, ie at R.

Now look at the across and up amounts to go from P to R and repeat these amounts in going from R to Q.

That will give you the coordinates of Q.


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