Question

find a and b if ax2+(2-b) xy +3y2 - 6bx+30y+6b is 0 if this is equation is of circle
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Answer 1

Answer:

a = 3 and b = 2

Step-by-step explanation:

We have been given the equation - $ax^{2}+(2-b)xy+3y^{2}-6bx+30y+6b=0$ and it is said that this equation represents a circle.

A General Second Degree Equation is of the form $Ax^{2}+2Hxy+By^{2}+2Gx+2Fy+C=0$

For a General Second Degree Equation to represent a circle, three conditions need to be fulfilled.

These conditions are:-

1. Co-efficient of $x^{2}$ and $y^{2}$ must be equal
2. The value of 'H' must be equal to zero.
3. The value for $delta \ \triangle=abc+2fgh-af^2-bg^2-ch^2 \neq 0$

Here, the co-effecient of $x^{2}$ is 'a' and the co-effectient of $y^{2}$ is 3.

They must be equal. So, a = 3

This equation has a term with the variable 'xy'. For it to be equal to zero, its co-effecient must be equal to zero.

So, (2-b) = 0
b = 2

Therefore, the value of a is 3 and b is 2.

The third point can be confirmed for further verification by substituing the values in the expression for delta.