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.

Good news: teeth all OK.

Bob

cooljackiec: I have added a new post to an old question of yours at

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x by 4

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

Bob

]]>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

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

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

]]>Let's check the steps.

What did you have for R ?

Bob

]]>It doesn't have to.

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

Bob

]]>(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

Bob

]]>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)."

]]>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.

Bob

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