急! F5 Equations of circles 8q2

13. Equation of the y - axis is x = 0. So put x = 0
a^2 + (y - b)^2 = (a - b)^2
(y - b)^2 = (a - b)^2 - a^2 = (a - b + a)(a - b - a) = - b(2a - b).
Since a > 0 and b < 0, that means - b(2a - b) is > 0
That is (y - b)^2 > 0, so there are 2 intersecting points.
15. (a)
x + y - 2 = 0, x = 2 - y. Sub. into the circle equation,
(2 - y)^2 + y^2 - 8(2 - y) + 8y + 22 = 0
4 - 4y + y^2 + y^2 - 16 + 8y + 8y + 22 = 0
2y^2 + 12y + 10 = 0
y^2 + 6y + 5 = 0
(y + 1)(y + 5) = 0
y = - 1 or - 5
So y - coordinate of the mid - point is [ (-1) + (-5)]/2 = - 3
Sub into the equation of the straight line, the x - coordinate of the mid - point is x = 2 - y = 5
So mid - point is (5, - 3)
(b) Center of circle is (4, - 4)
So perpendicular distance between center K and the line = distance between K and M = distance between (4, - 4) and ( 5, - 3).