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**cooljackiec****Member**- Registered: 2012-12-13
- Posts: 185

The graph of the equation

is a circle. Find the radius of the circle. is the radiusI see you have graph paper.

You must be plotting something

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**bobbym****bumpkin**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 109,535

Hi;

**In mathematics, you don't understand things. You just get used to them.****If it ain't broke, fix it until it is.**** Always satisfy the Prime Directive of getting the right answer above all else.**

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**bob bundy****Moderator**- Registered: 2010-06-20
- Posts: 7,948

hi cooljackiec,

I do these by completing the square:

So this is a circle with centre (1.5,-7) and radius 13.

If you got 269 it looks like a little arithmetic error to me.

Bob

Children are not defined by school ...........The Fonz

You cannot teach a man anything; you can only help him find it within himself..........Galileo Galilei

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**cooljackiec****Member**- Registered: 2012-12-13
- Posts: 185

The circles x^2 + y^2 = 4 and (x - 2)^2 + (y - 3)^2 = 7 intersect in two points A and B. Find the slope of \overline{AB}.

I see you have graph paper.

You must be plotting something

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**bobbym****bumpkin**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 109,535

Hi;

**In mathematics, you don't understand things. You just get used to them.****If it ain't broke, fix it until it is.**** Always satisfy the Prime Directive of getting the right answer above all else.**

**Online**

**cooljackiec****Member**- Registered: 2012-12-13
- Posts: 185

Find the center of the circle passing through the points (-1,0), (1,0), and (3,1). Express your answer in the form "(a,b)."

A line with slope 3 is 2 units away from the origin. Find the area of the triangle formed by this line and the coordinate axes.

I see you have graph paper.

You must be plotting something

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**bob bundy****Moderator**- Registered: 2010-06-20
- Posts: 7,948

hi cooljackiec,

I've made diagrams for both of these, below.

Q1

The perpendicular bisector of any chord goes through the centre of the circle.

So make two of these and see where they intersect.

Q2

Let the point where the sloping line cuts the circle radius 2 be (x,y)

The radius through this point must have slope -1/3 so that it is at right angles to the line with slope 3.

So y/x = -1/3 => 3y = -x

By pythag x^2 + y^2 = 4

Use these two equations to find x and y. **

Then you can get the equation of the sloping line.

Then you can get the intercepts with the axes and hence complete the question.

** There will be two solutions but you can use either as they both lead to the same size triangle.

Bob

Children are not defined by school ...........The Fonz

You cannot teach a man anything; you can only help him find it within himself..........Galileo Galilei

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**cooljackiec****Member**- Registered: 2012-12-13
- Posts: 185

why isnt the answer to the y/x 3????

I see you have graph paper.

You must be plotting something

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**bob bundy****Moderator**- Registered: 2010-06-20
- Posts: 7,948

I've re-made the diagram with the points labelled.

The line you want is AB. It has a slope of 3. A and B are the intercepts with the axes.

I've drawn a circle radius 2. AB touches the circle at X.

The difficulty is to work out where to place AB so that it does touch the circle at X.

To work this out I reasoned like this:

If AB is a tangent to the circle at X, then OX is a radius line and OX is perpendicular to AB

If AB has slope 3, then OX has slope -1/3, because the two slopes multiplied will always make -1.

(If you didn't know this try drawing a rectangle with sides 3 by 1 and another with sides 1 by 3. You'll see that the diagonal of 3 by 1 is at right angles to the diagonal of 1 by 3.)

So the line OX has equation y = -x/3 .

This will enable you to work out the coordinates of X. There are two possible answers; on opposite sides of the circle; both will lead to an answer for AB that works; one answer rotated 180 degrees will give the other answer.

Bob

Children are not defined by school ...........The Fonz

You cannot teach a man anything; you can only help him find it within himself..........Galileo Galilei

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**cooljackiec****Member**- Registered: 2012-12-13
- Posts: 185

oh.... so the area is 20/3 right??

Let P = (5,1), and let Q be the reflection of P in the line y = \frac{1}{2} x + 2. Find the coordinates of Q.

I see you have graph paper.

You must be plotting something

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**bob bundy****Moderator**- Registered: 2010-06-20
- Posts: 7,948

hi

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

You cannot teach a man anything; you can only help him find it within himself..........Galileo Galilei

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**cooljackiec****Member**- Registered: 2012-12-13
- Posts: 185

(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)."

I see you have graph paper.

You must be plotting something

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**bob bundy****Moderator**- Registered: 2010-06-20
- Posts: 7,948

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

Bob

You cannot teach a man anything; you can only help him find it within himself..........Galileo Galilei

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**cooljackiec****Member**- Registered: 2012-12-13
- Posts: 185

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

I see you have graph paper.

You must be plotting something

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**bob bundy****Moderator**- Registered: 2010-06-20
- Posts: 7,948

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

Bob

You cannot teach a man anything; you can only help him find it within himself..........Galileo Galilei

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**cooljackiec****Member**- Registered: 2012-12-13
- Posts: 185

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

I see you have graph paper.

You must be plotting something

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**bob bundy****Moderator**- Registered: 2010-06-20
- Posts: 7,948

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 ?

Bob

You cannot teach a man anything; you can only help him find it within himself..........Galileo Galilei

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**cooljackiec****Member**- Registered: 2012-12-13
- Posts: 185

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?

I see you have graph paper.

You must be plotting something

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**bob bundy****Moderator**- Registered: 2010-06-20
- Posts: 7,948

post 7 Q1

Bob

You cannot teach a man anything; you can only help him find it within himself..........Galileo Galilei

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**cooljackiec****Member**- Registered: 2012-12-13
- Posts: 185

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

I see you have graph paper.

You must be plotting something

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**bob bundy****Moderator**- Registered: 2010-06-20
- Posts: 7,948

17/4 is what I got too.

x by 4

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

Bob

You cannot teach a man anything; you can only help him find it within himself..........Galileo Galilei

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**cooljackiec****Member**- Registered: 2012-12-13
- Posts: 185

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

I see you have graph paper.

You must be plotting something

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**bobbym****bumpkin**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 109,535

Hi cooljackiec;

**In mathematics, you don't understand things. You just get used to them.****If it ain't broke, fix it until it is.**** Always satisfy the Prime Directive of getting the right answer above all else.**

**Online**

**bob bundy****Moderator**- Registered: 2010-06-20
- Posts: 7,948

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.

Good news: teeth all OK.

Bob

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

You cannot teach a man anything; you can only help him find it within himself..........Galileo Galilei

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**bob bundy****Moderator**- Registered: 2010-06-20
- Posts: 7,948

Above edited.

You cannot teach a man anything; you can only help him find it within himself..........Galileo Galilei

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