Circles Question

The circle C has centre (-1,6) and radius 2√5.
(a) find an equation for C.
The line y=3x-1 intersects C at the points A and B.
(b) Find the x-coordinates of A and B.
(c) Show that AB= 2√10

I'm not quite sure how to carry out all the right working out. Help is much appreciated :smile:

Reply 1

username3085646

OP

11

Oh and this one too please?

The circle C has centre (-2,2) and passes through the point (2,1).

(a) Find an equation for C.
(b) Show that the point with co-ordinates (-4,7) lies on C.
(c) Find an equation for the tangent to C at the point (-4, 7). Give your answer in the form ax+ by+ c=0, where a, b and c are integers.

Reply 2

RogerOxon

21

Original post by bluepotato

The circle C has centre (-1,6) and radius 2√5.
(a) find an equation for C.
The line y=3x-1 intersects C at the points A and B.
(b) Find the x-coordinates of A and B.
(c) Show that AB= 2√10

I'm not quite sure how to carry out all the right working out. Help is much appreciated :smile:

The general equation of a circle is:

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

It's basically Pythagoras' theorem saying that the distance from the centre to any point on the circumference is

r

.

If you substitute for

y

from your line equation, then you'll get a quadratic. Solve that and you have the

x

coordinates of the points of intersection (if any). Use the line equation to get the

y

coordinates of those points.

(edited 6 years ago)

Reply 3

RogerOxon

21

Original post by bluepotato

Oh and this one too please?

The circle C has centre (-2,2) and passes through the point (2,1).

(a) Find an equation for C.

See above.

(b) Show that the point with co-ordinates (-4,7) lies on C.

Any point on C must satisfy the equation in (a), so see if it does.

(c) Find an equation for the tangent to C at the point (-4, 7). Give your answer in the form ax+ by+ c=0, where a, b and c are integers.

What is the gradient of the line from the centre of C to that point?
What is the gradient of the line perpendicular to that one?
Use

y=mx+c

to obtain the line with the required gradient and point. Rearrange it to the form requested.

(edited 6 years ago)

OP

Thanks a bunch! :h:

OP

Does anyone reckon they could do the first question on here?

Reply 6

RogerOxon

21

Original post by bluepotato

Does anyone reckon they could do the first question on here?

Yes. We don't give full solutions, just point you in the right direction.

If you're having problems solving it, please post what you've done so far.

Reply 7

username3085646

OP

11

Right I've found the equation (part a) which is (x+1)^2 + (3x-7)^2=20

Not sure how I should go ahead for part b and c

Should I substitute the co-ordinates for part b?

Reply 8

username3085646

OP

11

Or do the quadratic equation like : x2-31x+30 = 0 and solve it?

Reply 9

RogerOxon

21

Original post by bluepotato

Right I've found the equation (part a) which is (x+1)^2 + (3x-7)^2=20

Original post by bluepotato

Or do the quadratic equation like : x2-31x+30 = 0 and solve it?

Expand the first equation. It doesn't give the quadratic that you quoted. You should be able to remove a factor of 10 to give a quadratic with some obvious roots. Can you show your working?