Question

The abscissa of two points A and B are the roots of the equation${x^2} + 2ax - {b^2} = 0$and their ordinates are the root of the equation${x^2} + 2px - {q^2} = 0$. The equation of the circle with AB as diameter is
A. ${x^2} + {y^2} + 2ax + 2py + {b^2} + {q^2} = 0$
B. ${x^2} + {y^2} - 2ax - 2py - {b^2} - {q^2} = 0$
C. ${x^2} + {y^2} + 2ax + 2py - {b^2} - {q^2} = 0$
D. None of these

Answer 1

Hint: We can take A and B as $\left( {{x_1},{y_1}} \right)$and$\left( {{x_2},{y_2}} \right)$. Now we have ${x_1}$and${x_2}$as root of the 1st equation. Then we can find the sum of the roots and product of the roots. Similarly, we can find the product and sum of y coordinates. Then we can take the diameter form formula of circle which is given by\[\left( {x-{x_1}} \right)\left( {x-{x_2}} \right) + \left( {y-{y_1}} \right)\left( {y-{y_2}} \right) = 0\]. We can simplify it substitute the sum and product of the coordinates to get the required equation of the circle.

Complete step by step answer:

Let A$\left( {{x_1},{y_1}} \right)$ and B$\left( {{x_2},{y_2}} \right)$be the 2 points.
We have the equation${x^2} + 2ax - {b^2} = 0$. According to the question, its roots are${x_1}$and${x_2}$.
For a quadratic equation of the form$a{x^2} + bx + c = 0$, sum of the root is given by $ - \dfrac{b}{a}$ and product of root is$\dfrac{c}{a}$.
Here the sum of roots is given by,
${x_1} + {x_2} = \dfrac{{ - 2a}}{1} = - 2a$ … (1)
And product of roots is
\[{x_1}{x_2} = \dfrac{{ - {b^2}}}{1} = - {b^2}\]… (2)
We have the equation${x^2} + 2px - {q^2} = 0$. According to the question, its roots are\[{y_1}\]and${y_2}$.
Here the sum of roots is
${y_1} + {y_2} = \dfrac{{ - 2p}}{1} = - 2p$ … (3)
And product of roots is
\[{y_1}{y_2} = \dfrac{{ - {q^2}}}{1} = - {q^2}\]… (4)
We know that, if A$\left( {{x_1},{y_1}} \right)$ and B$\left( {{x_2},{y_2}} \right)$ are the endpoints of the diameter, then the equation of the circle is given by,
\[\left( {x-{x_1}} \right)\left( {x-{x_2}} \right) + \left( {y-{y_1}} \right)\left( {y-{y_2}} \right) = 0\]
On taking the product, we get,
\[{x^2}-{x_1}x-{x_2}x + {x_1}{x_2} + {y^2}-{y_1}y-{y_2}y + {y_1}{y_2} = 0\]
On rearranging, we get,
\[ \Rightarrow {x^2} + {y^2}-\left( {{x_1} + {x_2}} \right)x-\left( {{y_1} + {y_2}} \right)y + {x_1}{x_2} + {y_1}{y_2} = 0\]
Substituting equations (1), (2), (3) and (4), we get,
\[ \Rightarrow {x^2} + {y^2}-\left( { - 2a} \right)x-\left( { - 2p} \right)y + - {b^2} + - {q^2} = 0\]
$ \Rightarrow {x^2} + {y^2} + 2ax + 2py - {b^2} - {q^2} = 0$
Therefore the required equation is ${x^2} + {y^2} + 2ax + 2py - {b^2} - {q^2} = 0$
So the correct answer is option C.

Note: Alternate approach to solve this problem is,
We have the equation${x^2} + 2ax - {b^2} = 0$. According to the question, its roots are${x_1}$and${x_2}$. Then $\left( {x - {x_1}} \right)$and $\left( {x - {x_2}} \right)$are factors of the equation.
So we can write, $\left( {x - {x_1}} \right)\left( {x - {x_2}} \right) = 0$
$ \Rightarrow {x^2} + 2ax - {b^2} = \left( {x - {x_1}} \right)\left( {x - {x_2}} \right)$… (a)
We have the equation${x^2} + 2px - {q^2} = 0$.we can change the variable to y.
$ \Rightarrow {y^2} + 2py - {q^2} = 0$
According to the question, its roots are\[{y_1}\]and${y_2}$. Then $\left( {y - {y_1}} \right)$and $\left( {y - {y_2}} \right)$are factors of the equation.
So we can write, $\left( {y - {y_1}} \right)\left( {y - {y_2}} \right) = 0$
$ \Rightarrow {y^2} + 2py - {q^2} = \left( {y - {y_1}} \right)\left( {y - {y_2}} \right)$… (b)
We know that, if A$\left( {{x_1},{y_1}} \right)$ and B$\left( {{x_2},{y_2}} \right)$ are the endpoints of the diameter, then the equation of the circle is given by,
\[\left( {x-{x_1}} \right)\left( {x-{x_2}} \right) + \left( {y-{y_1}} \right)\left( {y-{y_2}} \right) = 0\]
Substituting, (a) and (b), we get,
${x^2} + 2ax - {b^2} + {y^2} + 2py - {q^2} = 0$
On rearranging, we get,
$ \Rightarrow {x^2} + {y^2} + 2ax + 2py - {b^2} - {q^2} = 0$
Therefore the required equation is ${x^2} + {y^2} + 2ax + 2py - {b^2} - {q^2} = 0$