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1. https://a44694f152ee562e38a9454969d2...%20Edexcel.pdf

question 5 a i have no idea how to do this help pls

Zacken
2. (Original post by thefatone)
https://a44694f152ee562e38a9454969d2...%20Edexcel.pdf

question 5 a i have no idea how to do this help pls

Zacken
GCSE theorem, angle in a semi-cricle is a right-angle.

Here PR is a diameter so PQ and QR form an angle in a semi-circle, making an angle of 90 degrees, hence PQ and QR are perpendicular. What can you then say about their gradients? More specficially, their product of gradients. That'll give you an equation in a which you can then solve.
3. Calculate the gradient of PQ and then as QR is perpendicular the the gradient of QR will be -3/2

Then work out the equation of the line QR by susbstituting in the coordinate of Q to get the equation

Plug in y = 4 and calculate a to be 13
4. (Original post by Zacken)
GCSE theorem, angle in a semi-cricle is a right-angle.

Here PR is a diameter so PQ and QR form an angle in a semi-circle, making an angle of 90 degrees, hence PQ and QR are perpendicular. What can you then say about their gradients? More specficially, their product of gradients. That'll give you an equation in a which you can then solve.
i don't understand

......

oh no i see ok right, i didn't see they were perpendicular right now i see thanks
5. (Original post by thefatone)
i don't understand

......

oh no i see ok right, i didn't see they were perpendicular right now i see thanks
6. (Original post by Zacken)
we're getting there slowly all these silly things which i don't see -.-
7. a ) Diameter^2 : (a + 3)^2 + (4 - 2)^2 = a^2 + 6a + 9 + 4 = a^2 + 6a + 13

(a - 9)^2 + ( 4 - 10)^2 = a^2 - 18a + 81 + 36 => a^2 - 18a + 117

a^2 + 6a + 13 = 208 + a^2 - 18a + 117

=> 24a = 325 - 13 = 312

a = 13

b)Centre x = a + 3 / 2 = 8

Centre y = 4 - 2 / 2 = 1

Diameter^2 = 13^2 + 6 * 13 + 13 = 169 + 78 + 13 = 91 + 169 = 260

Diameter = 16.12

(x - 8)^2 + (y - 1)^2 = 65

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