The line joining the points A (0,5) and B(4,1) is a tangent to a circle whose centre is at point (5,4)
(a) find the equation of line AB
(b)find the equation of line through C which is perpendicular to AB
(c)find the co-ordinates of the point of contact of the line AB with the circle
(d) find the equation of the circle



THANKS TO ANY ONE WHO CAN HELP ME OUT!

(a)
(y - 5)/(x - 0) = (1 - 5)/(4 - 0) and then solve...

(b)
since AB (tangent to the circle) is perpendicular to CO (O is centre of circle and C is the point of contact between circle and AB), the product of the two slopes = -1
first find the slope of AB from the answer in (a), then use it to find the slope of CO.
then use the slope of CO and coordinates of O(5,4) to work out its equation.

(c) solve the two simultaneous equations obtained from (a) and (b)

(d) find the distance between O(5,4) and the answer from (c), which is the point of intersection between AB and CO. this distance is radius of circle. finally, use this radius r and O(5,4) to work out the equation of circle, ie:

(y - 4)^2 + (x - 5)^2 = r^2
and present answer in the form of Ax^2 + By^2 + Cx + Dy + constant = 0

__________________
GeoLover

You know for part a how wud u slove from there? thanks.

(a)


y - 5 = (-1)(x) = -x
y = -x + 5
(equivalant to y = mx + c, hence slope m = -1 and y-intercept c = +5)

(b)
m_AB = slope of AB = -1
m_CO = slope of CO
since m_AB x m_CO = -1
m_CO = 1
hence
(y - 4) = (x - 5)
hence we get y = x - 1

(c)
y = -x + 5 .....(1)
y = x - 1 .....(2)
hence -x + 5 = x - 1
x = 3
and y = 2
hence the coordinates of point of contact is C(3,2)

(d)
distance between C(3,2) and O(5,4) = r
= [(4 - 2)^2 + (5 - 3)^2]^0.5
= 8^0.5
finally the equation of circle:
(y - 4)^2 + (x - 5)^2 = r^2
(y^2 - 8y + 16) + (x^2 -10x + 25) = (8^0.5)^2
x^2 + y^2 - 10x - 8y + 16 + 25 - 8 = 0
x^2 + y^2 - 10x - 8y + 33 = 0

__________________
GeoLover