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# Circles Question watch

1. 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
2. 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.
3. (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
The general equation of a circle is:

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

If you substitute for from your line equation, then you'll get a quadratic. Solve that and you have the coordinates of the points of intersection (if any). Use the line equation to get the coordinates of those points.
4. (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 to obtain the line with the required gradient and point. Rearrange it to the form requested.
5. Thanks a bunch!
6. Does anyone reckon they could do the first question on here?
7. (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.
8. 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?
9. Or do the quadratic equation like : x2-31x+30 = 0 and solve it?
10. (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?
11. Apologies. It's 10x^2 - 33x + 50 =0
12. Now I solve for x?
13. (Original post by bluepotato)
Apologies. It's 10x^2 - 33x + 50 =0
Please post your working for that - the 33 and 50 are wrong. You seem to have forgotten the 20 on the RHS.
14. 10x^2-40x+30 =0
15. (Original post by bluepotato)
10x^2- 50 x +50 =0

Now collect the terms and move the 20 over. You should then be able to divide through by 10 and factorise.
16. Yup gotcha. Omds I made such silly mistakes! How do I solve for part c then where we need to show that AB=2root10
17. (Original post by bluepotato)
10x^2-40x+30 =0
That's it. Divide through by 10 and factorise. Then substitute into the line equation to get the points.
18. Btw my x values were x=3 or x=1 which I'm presuming is right
19. (Original post by bluepotato)
Yup gotcha. Omds I made such silly mistakes! How do I solve for part c then where we need to show that AB=2root10
If you have the coordinates of A and B, then calculate the x and y differences and use Pythagoras to get the distance.
20. (Original post by bluepotato)
Btw my x values were x=3 or x=1 which I'm presuming is right
Yes, that's right.

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