VENNA RAMANJANEYA REDDY
Last Activity: 10 Years ago
Hi,
Let given points P(x1,y1) = (3,7)
Q(x2,y2) = (5,5)
Assume center of the circle is C(a,b) and midpoint of the chord PQ is M(x3,y3).
Any two points on the circle forms a chord. If we draw a line from midpoint point of the chord and center of the circle, the line will be perpendicular to the chord.
i.e CM line is perpendicular to the PQ line.
Midpoint point of the chord PQ, M(x3,y3) = ( (x1+ x2)/2, (y1+y2)/2 )
= ( (3+5)/2, (7+5)/2 )
= (8/2, 12/2 )
M(x3,y3) = (4,6)
CM line is perpendicular to the PQ line so , Slope of the line CM * slope of line PQ = – 1 ----(i)
Slope of line PQ = (y2-y1)/(x2-x1)
= (5-7) / (5-3)
= -2/2
Therefore Slope of line PQ = -1
From (i), Slope of line CM = - 1 / slope of line PQ = -1 / -1 = 1
Equation of line CM passing through M with slope 1 is
y-y3 = m (x-x3)
y-6= 1 ( x-4)
y-6 = x-4
Equation of line CM x-y = - 2 -----------------------(ii)
Given equation of line which passes through center is x-4y = 1 --------------(iii)
The intersection if line (ii) and (iii) is the center of the circle.
By solving (ii) & (iii), we get x= -3, y = -1
So Center of the circle C(a,b) = (-3, -1).
Radius of the circle is distance from center C to point of the circle P.
r = distance between CP = sqrt ( 36 + 64) = sqrt (100) = 10
The equation of the circle is (x-a)2 + (y-b)2 = r2
(x+3) square + (y+1) square = 100
