If a circle passes through points (0,a), (a,0),(0,b) then find its centre and circuits
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Answer:
(a/2 , b/2)
Step-by-step explanation:
The general equation of the circle is given by
x² + y² + 2gx + 2fy + c = 0
This equation always represents a circle whose centre is (-g, -f) and
radius = √( g2+f2−c), where g, f and c are three constants
Given that,
the point (0,0), put this in the equation of the circle
0² + 0² + 2(0)g + 2(0)f + c = 0
c = 0
the point (a,0), put this in the equation of the circle
a²+ 0² +2ga + c = 0
a²+ 0² +2ga + 0 = 0
a²+2ag = 0
a(a+2g) = 0
g = − a/2
Given the point (0,b) put this in the equation of the circle
0² + b² + 2fb + c = 0
b²+2fb = 0
b(b+2f) = 0
f = −b/2
We know that the centre of the circle = (−g,−f)
= (a/2 , b/2)
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