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# As core 1 Equation of a circle- Sixth form watch

1. Okk i dont see where ive gone wrong, if someone could help?

the question is
1). P(6,7) is a point on a circle with its centre at (-3,1), find the equation of the normal to circle at P

So i found the gradient first and i got 2/3, i then put it into the forumla

y-7=2/3(x-6) i then x3 to get rid of the fraction and got 3(y-7)=2(x-6) and expanded it to get 3y-21=2x-12 and rearranged it to make y the subject and got the answer:

y=2/3x +9 but the asnwer book said it was y=2/3x+3

what did i do wrong?
2. 3y-21 = 2x-12
3y = 2x +9
y= 2/3 x + 3
3. Right, well you know the general formula of a circle is:

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

So to begin with you have:

(x+3)^2+(y-1)^2=r^2

Then you need to find the distance between the two points, giving the radius of the circle. You just need phythagoras here:

Root: (x2-x1)^2+(y2-y1)^2

And as that is r^2, you don't need to root that at all.

(x+3)^2+(y-1)^2=127

I think, correct me if am wrong
4. (Original post by AspiringGenius)
Right, well you know the general formula of a circle is:

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

So to begin with you have:

(x+3)^2+(y-1)^2=r^2

Then you need to find the distance between the two points, giving the radius of the circle. You just need phythagoras here:

Root: (x2-x1)^2+(y2-y1)^2

And as that is r^2, you don't need to root that at all.

(x+3)^2+(y-1)^2=127

I think, correct me if am wrong
Almost, 81 + 36 = 117 not 127

Also latex helps a lot in answering questions - I was a bit overwhelmed by it at first but it is really easy to use when you get used to it.

EDIT: But more importantly I just realised you didn't answer the question at all It was asking the equation of the normal to the tangent at point P.
5. (Original post by hassi94)
Almost, 81 + 36 = 117 not 127

Also latex helps a lot in answering questions - I was a bit overwhelmed by it at first but it is really easy to use when you get used to it.
Damn. I hate you basic arithmetic!
6. (Original post by Gary)
Okk i dont see where ive gone wrong, if someone could help?

the question is
1). P(6,7) is a point on a circle with its centre at (-3,1), find the equation of the normal to circle at P

So i found the gradient first and i got 2/3, i then put it into the forumla

y-7=2/3(x-6) i then x3 to get rid of the fraction and got 3(y-7)=2(x-6) and expanded it to get 3y-21=2x-12 and rearranged it to make y the subject and got the answer:

y=2/3x +9 but the asnwer book said it was y=2/3x+3

what did i do wrong?
Once you have the gradient of 2/3 you have:

Hence,
7. is it just me or is the gradient of a normal -1/m

and the gradient here is 2/3 => gradient of normal is -3/2 ??? or am i just being silly
8. (Original post by meraphox)
is it just me or is the gradient of a normal -1/m

and the gradient here is 2/3 =&gt; gradient of normal is -3/2 ??? or am i just being silly
The gradient of 2/3 was worked out as the line from the centre to point P, which is the same as the gradient of the normal (and exactly the same line for that matter).

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