amaths - circle

Family of circle is
(3x^2 + 3y^2 +13x - y -40) + k(x^2 + y^2 + 4x +2y -20) = 0.
Now (0,0) is on the circle, therefore put x=0 and y= 0 we get
-40 + k(-20) = 0. Therefore, k = -2. So the required circle is
(a)3x^2 + 3y^2 + 13x - y - 40 - 2x^2 - 2y^2 - 8x - 4y + 40 = 0
x^2 + y^2 + 5x -5y = 0.........................(1)
(b)Equation of the intersecting chord of C1 and C2 is
(13 - 4 x 3) x + (-1 -3 x 2)y +(-40 +20x 3) = 0
x - 7y + 20 = 0.......................(2), which is the equation of PQ.
From (1), centre of circle is (-5/2, 5/2). Substitute this into (2)
(-5/2) - 7(5/2) + 20 = -5/2 - 35/2 + 20 = -40/2 + 20 = 0.
That means centre of the required circle is on PQ, so PQ is the diameter.
Therefore, angle of the origin with PQ is a right angle by property of angle in semi-circle.
(c) Centre of C1 is (-2,-1). Substitute this into the required circle, we get
(-2)^2 + (-1)^2 + 5(-2) -5(-1) = 4 + 1 - 10 + 5 = 0. That means centre of circle C1 is also on the required circle. So same as the origin, angle substened by PQ and centre of C1 is also a right angle.

(a) The family of circles passing through P and Q has an equation of:
k(x2 + y2 + 4x + 2y - 20) + (3x2 + 3y2 + 13x - y - 40) = 0
To give a circle passing through the origin, we sub x = y = 0 into the equation, then
k(- 20) + (- 40) = 0
k = -2
So the equation of the requred circle is:
-2(x2 + y2 + 4x + 2y - 20) + (3x2 + 3y2 + 13x - y - 40) = 0
-2x2 - 2y2 - 8x - 4y + 40 + 3x2 + 3y2 + 13x - y - 40 = 0
x2 + y2 + 5x - 5y = 0
(b) Observe the mid-point of PQ lies on family of the circle when the family gives a straight line, i.e. k = -3 and the equation is:
-3(x2 + y2 + 4x + 2y - 20) + (3x2 + 3y2 + 13x - y - 40) = 0
-12x - 6y + 60 + 13x - y - 40 = 0
x - 7y + 20 = 0
And also, the centre of the circle in (a) is (-5/2, 5/2) which also lies on the straight line found.
So, PQ is in fact the diameter of the circle in (a) and hence ∠POQ is a right angle for the reason of angle in semi circle.
FInally OP is perpendicular to OQ.
(c) Let L be the line joining the centres of C1 and C2. Then L passes through the mid point of PQ, i.e. centre of the circle in (a) and also L is perpendicular to PQ.
So radius of the circle in (a) = 5/√2
Radius of C1 = 5
Now, let the required angle be θ, then we have:
sin (θ/2) = (5/√2)/5 = 1/√2
θ/2 = 45
θ = 90