if the circle x^2+y^2+ax+by-12=0has the center at 2,3 then find a,b and the radius of the circle
Answer (1)
9th Oct, 2020
Hi,
The general equation of a circle is ax^2+2hxy+by^2+2gx+2fy+c=0.
Here a=1 b=1 h=0
So equation is x^2 + y^2 + 2gx + 2fy + c = 0
In such equations,
centre = (-g,-f)
radius = √(g^2 + f^2 - c )
Equation of the circle here is x^2 + y^2 + ax + by – 12 = 0
Centre = (-a/2, -b/2) = (2, 3)
−a/2 = 2, − b/2 = 3
a = – 4, b = – 6
g = –2, f = –3, c = –12
radius = √(g^2 + f^2 - c) = √(4 + 9 + 12) = 5
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