maths question, circles etc.?

Question

this one is tough.

The points p and q have coordinates (-2,6) and (4, -1) respectively.
Given that PQ is a diameter of circle C

i) Find the coordinates of the centre C

ii) sgiw that C has the equation

X^2 + Y^2 - 2x -5y - 14 =0

The point R has coordinates (2,7)

iii) Show that R lies on C and hence state the size of

Answer 1

The centre is the midpoint of PQ so C(1, 5/2)

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The standard equation of a circle centre (a,b) radius r is

(x - a)^2 + (y-b)^2 = r^2 = d^2/4

now d = distance PQ = sqrt( 6^2 + 7^2) = sqrt85 by Pythagoras

so (x - 1)^2 + (y- 5/2)^2 = 85/4

or x^2 - 2x +1 + y^2 -5y +25/4 = 85/4

or x^2 - 2x- 5y -14 = 0

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To show that R lies on the circle just substitute the point into the LHS and show it comes to zero


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Last part is some old GCSE work which says that the angle at the circumference in a semicircle is a right angle

Answer 2

(i) Centre (1,5/2)

(ii)
r² = (4 - 1)² + (- 1 - 5/2)²
r² = 9 + 49 / 4
r² = 36 / 4 + 49 / 4 = 85/4
(x - 1)² + (y - 5/2)² = 85/4
x² - 2x + 1 + y² - 5y + 25/4 = 85/4
x² + y² - 2x - 5y + 1 = 15
x² + y² - 2x - 5y - 14 = 0

(iii) Test for (2,7) :-
4 + 49 - 4 - 35 - 14 = 53 - 53 = 0 as required.
R lies on circumference and PQ is a diameter therefore angle PQR is a right angle.

Answer 3

i) Find the coordinates of the centre

centre= (h,k)= mid point of (-2,6) and (4, -1)

ii) sgiw that C has the equation

X^2 + Y^2 - 2x -5y - 14 =0

radius = distance from (4, -1) to (h,k)

equation = (x-h)^2+(y-k)^2=r^2......and remove brackets

iii) Show that R lies on C

plug (2,7) into C

Answer 4

Ooh me brain hurts!