find the values of a = b if the equation (a-4) x^2+by^2+(b-3)xy+4x+4y-1=0 represents a circle
Answers
Answer:
For a second degree equation in x and y
ax^2 + 2hxy + by^2 +2gx + 2fy + c= 0
to represent a circle,
a = b and h= 0
For the given equation a=b and a+b-4 = 0 => a=b= 2
Hence the given equation reduces to
2x^2 + 2y^2 - 2x - 2y - 20 = 0
i.e., x^2 + y^2 - x - y - 10 = 0.
This can be written as
(x- 1/2)^2 + (y - 1/2)^2 = {√(21/2)}^2
This is of the form (x-h)^2 + (y-k)^2 = r^2
Hence the radius of the given circle is √(21/2).
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