急!F5 Equations of Circles 8q15

(a)

centre M: (-3, 5)
radius = sqrt(3^2 + 5^2 - 24) = sqrt(10)
the distance between A and centre M
= sqrt{[(-4) - (-3)]^2 + (6 - 5)^2} = sqrt(2) < sqrt(10) = radius
therefore A lies inside C

the slope of AM = (6 - 5) / [(-4) - (-3)] = -1
the slope of require chord = -1 / -1 = 1
therefore the equation is y - 6 = 1(x + 4) => x - y + 10 = 0

(b)

y = x + 10 (from a)
substitute y = x + 10 into x^2 + y^2 + 6x - 10y + 24 = 0
x^2 + (x + 10)^2 + 6x - 10(x + 10) + 24 = 0
2x^2 + 16x + 24 = 0 => x = -2, -6
when x = -2, y = 8; when x = -6, y = 4
therefore the coordinates of P and Q are (-2, 8) and (-6, 4)

(c)

slope of PM = (8 - 5) / [(-2) - (-3)] = 3
slope of PR = -1/3
the equation of PR is y - 8 = (-1/3)(x + 2) => x + 3y - 22 = 0

slope of QM = (4 - 5) / [(-6) - (-3)] = 1/3
slope of PR = -1 / (1/3) = -3
the equation of PR is y - 4 = -3(x + 6) => 3x + y + 14 = 0

the coordinates of R are
the cooradinates ot the point of intersection of PR and QR. i.e. (-8. 10)

(d)(i)

PR = sqrt{[(-2) - (-8)]^2 + (8 - 10)^2} = sqrt(40) = 2sqrt(10)
area of triangle MPR = (1/2)[2sqrt(10)]sqrt(10) = 10
area of quadrilateral PRQM = 10 x 2 = 20

(d)(ii)

the diameter of require circle is MR
centre of require circle is ([(-3) + (-8)]/2, (5 + 10)/2), i.e. (-5.5, 7.5)
radius of require circle = (1/2)MR
= (1/2)sqrt{[(-3) - (-8)]^2 + (5 - 10)^2)} = (1/2)sqrt(50) = (5/2)sqrt(2)
thus the equation of require circle is
(x + 5.5)^2 + (y - 7.5)^2 = [(5/2)sqrt(2)]^2
=> x^2 + 11x + 30.25 + y^2 - 15y + 56.25 = 12.5
=> x^2 +y^2 + 11x - 15y + 74 = 0