help

a circle passes through A(3,1) b(8,2) C(2,6)

A) Find the point of intersection of the perpendicular bisectors of AB and BC

do I just find the mid points and gradients of AB and BC and the take the negative reciprocal of the gradient and form an equation for AB and AC using y-1=m(x-x1) and solve them simultaneously to get the points of intersection? thanks :smile:

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yes

Original post by L'Evil Fish

yes

thanks

Reply 3

the bear

well since these are points on a circle the bisectors will be diameters, so it is asking you to find the center of the circle ?

Reply 4

madfish

Original post by L'Evil Fish

yes

hey fish how do I do this? :smile:

a circle touches the positive x-axis and y axes and its centre lies on the line y=3x-4

find an equation of the circle

Reply 5

Noble.

Original post by madfish

hey fish how do I do this? :smile:

a circle touches the positive x-axis and y axes and its centre lies on the line y=3x-4

find an equation of the circle

What is the general equation of a circle?

(edited 11 years ago)

Reply 6

madfish

Original post by Noble.

What is the general equation/form of a circle?

(x-a)^2+(y-b)^2=r^2

Reply 7

Noble.

Original post by madfish

(x-a)^2+(y-b)^2=r^2

Ok, what do you know about the point

(a,b)

. Also, how can you use the fact it touches both axes?

Reply 8

madfish

Original post by Noble.

Ok, what do you know about the point

(a,b)

. Also, how can you use the fact it touches both axes?

(a,b) lies on the line y=3x -4

Reply 9

Noble.

Original post by madfish

(a,b) lies on the line y=3x -4

Yep. Also, given that it touches the axes - what does this mean?

Reply 10

madfish

Original post by Noble.

Yep. Also, given that it touches the axes - what does this mean?

um, I am not too sure. :/

Reply 11

madfish

Original post by Noble.

Yep. Also, given that it touches the axes - what does this mean?

something about the radius?

Reply 12

Noble.

Original post by madfish

um, I am not too sure. :/

Well, if it touches both axes we know that there is a solution when

x=0

and when

y=0

.

This is much more easily visualised if you draw the line

y=3x-4

and you essentially are looking for a point on the line where you can draw a circle and have it just touch both axes - this point will be unique. However, for the equation

y=x

you should be able to see there are an infinite number of solutions. For example, if the centre was

(1,1)

and you had the radius as

1

it would satisfy the problem, and in fact any centre of

(p,p)

with radius

p

works - there are infinitely many solutions. However, because this line is not going through the origin we're looking for a point on the line

y=3x-4

which is at equal distances from the y-axis and the x-axis. Again, if you don't understand this, you need to draw it. Clearly, the centre

(5,11)

is on the line

y=3x-4

- but can we form a circle which touches both the x-axis and y-axis?

(edited 11 years ago)

Reply 13

madfish

Original post by Noble.

Well, if it touches both axes we know that there is a solution when

x=0

and when

y=0

.

This is much more easily visualised if you draw the line

y=3x-4

and you essentially are looking for a point on the line where you can draw a circle and have it just touch both axes - this point will be unique. However, for the equation

y=x

you should be able to see there are an infinite number of solutions. For example, if the centre was

(1,1)

and you had the radius as

1

it would satisfy the problem, and in fact any centre of

(p,p)

with radius

p

works - there are infinitely many solutions. However, because this line is not going through the origin we're looking for a point on the line

y=3x-4

which is at equal distances from the y-axis and the x-axis. Again, if you don't understand this, you need to draw it. Clearly, the centre

(5,11)

is on the line

y=3x-4

- but can we form a circle which touches both the x-axis and y-axis?

okay I am so lost

how did you deduce the centre?

Reply 14

Noble.

Original post by madfish

okay I am so lost

how did you deduce the centre?

What do you mean how do I deduce the centre? The equation of a circle:

(x-a)^2 + (x-b)^2 = r^2

has centre

(a,b)

and radius

r

We know the centre lies on the line

y=3x-4

which implies that

b = 3a-4

Reply 15

madfish

Original post by Noble.

What do you mean how do I deduce the centre? The equation of a circle:

(x-a)^2 + (x-b)^2 = r^2

has centre

(a,b)

and radius

r

We know the centre lies on the line

y=3x-4

which implies that

b = 3a-4

okay where do we go from here?: :colondollar:

Reply 16

joostan

Original post by madfish

okay I am so lost

how did you deduce the centre?

He didn't. He simply gave an example of a point that lies on the line y = 3x -4.

Reply 17

Noble.

Original post by madfish

okay where do we go from here?: :colondollar:

Honestly, there's no point in me spoon feeding you all the way to the answer, it doesn't help you.

You know that the circle 'touches' the x-axis and the y-axis - do you know what this means?