Circle question C2

last one i swear
part c espically

The circle C, with centre A, has equation

x^2 + y^2 - 6x + 4y - 12 = 0.

(a) Find the coordinates of A.

(b) Show that the radius of C is 5.

The points P, Q and R lie on C. The length of PQ is 10 and the length of PR is 3.

(c) Find the length of QR, giving your answer to 1 decimal place.

Reply 1

CalculusMan

a) Complete the square to get into (x - a)^2 + (y - b)^2 = r^2. A is (a, b)
b) Square root r^2 in eq. to get the radius.
c) Pythagoras.

Reply 2

danniella

Hello!

First part:
x^2 + y^2 -6x + 4y - 12 = 0
x^2 - 6x + y^2 + 4y = 12
(x-3)^2 - 9 + (y+2)^2 - 4 = 12 (completing the square)
(x-3)^2 + (y+2)^2 = 25
(x-3)^2 + (y+2)^2 = 5^2

so the circle has centre (3, -2) and radius 5.

Second part: PQ is exactly the length of 2 radii of the circle, and as P and Q lie on the circle, PQ is a diameter of the circle. R is also on the circumference of the circle, so PRQ is a right-angled triangle (angle R=90 degrees) as it's a triangle in a semicircle. Um. As R is the angle in a semicircle. Or something. Then PQ is the hypotenuse of the right angled triangle PRQ

So (PQ)^2 = (PR)^2 + (QR)^2
100 = (PR)^2 + 9
91 = (PR)^2
so
PR = root91 = 9.539 units to 3dp

love danniella