*Question*

# The abscissa of 2 points A and B are the roots of the equation \(x^2 + 2x – a^2 = 0\) and the ordinate are the roots of the equation \(y^2 + 4y – b^2 = 0\). Find the equation of the circle with AB as a diameter.

*Solution:*

Let roots of the equation \(x^2 + 2x – a^2 = 0\) are \(x_1 \; and\; x_2\)

and roots of the equation \(y^2 + 4y – b^2 = 0\) are \(y_1 \; and \; y_2\)

Therefore \(x_1 + x_2 = \frac{-2}{1} \;and \;x_1x_2 = \frac{-a^2}{1}\)

Similarly \(y_1 + y_2 = \frac{-4}{1} \;and\; y_1y_2 = \frac{-b^2}{1}\)

Hence abscissa and ordinate of A are \(x_1\; and \; y_1\) respectively.

Similarly for B are \(x_2\;and\;y_2\).

we know the diametric form of circle is \((y – y_1)(y – y_2) = -(x – x_1)(x – x_2)\)

\(\Rightarrow x^2 – (x_1 + x_2)x + x_1x_2 + y^2 – (y_1 + y_2)y + y_1y_2 = 0\)

\(\Rightarrow x^2 -(-2)x + (-a^2) + y^2 – (-4)y + (-b^2) = 0 \)

Hence Equation required = \(x^2 + y^2 + 2x + 4y – (a^2 + b^2) = 0 \)

**Here is the Second and Easy Method:**

If roots of two quadratic equations say \(ap^2 + bp + c = 0\) and \(dq^2 + eq + f = 0\), represent endpoints(one equation their abscissa and other ordinates) of a diameter of the circle, then the equation of the circle will be the summation of both the equations after making their second-degree term coefficient = 1.

That is C : \(p^2 + \frac{b}{a}p + \frac{c}{a} + q^2 + \frac{e}{d}q + \frac{f}{d} = 0\)