Question

The abscissa and ordinate of the points $A$ and $B$ are the roots of the equation ${x^2} + 2ax + b = 0$ and ${x^2} + 2cx + d = 0$ respectively, then the equation of the circle with AB as diameter is
A. ${x^2} + {y^2} + 2ax + 2cy + b + d = 0$
B. ${x^2} + {y^2} - 2ax - 2cy - b - d = 0$
C. ${x^2} + {y^2} - 2ax - 2cy + b + d = 0$
D. ${x^2} + {y^2} + 2ax + 2cy - b - d = 0$

Answer 1

In this question we have been given two quadratic equations. So we will find the roots of both the equations with the quadratic formula. We will assume the coordinates are $({x_1},{y_1})({x_2},{y_2})$. After this we will apply the equation of the circle: $(x - {x_1})(x - {x_2})(y - {y_1})(y - {y_2})$.

Complete step by step answer:
Let us assume the coordinates are $({x_1},{y_1})({x_2},{y_2})$. We will take the first equation i.e.
${x^2} + 2ax + b = 0$
We will use the quadratic formula which is
$\dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}$
In this equation we have
$b = 2a,c = b,a = 1$

Now by applying the formula we can write
$x = \dfrac{{ - 2a \pm \sqrt {{{\left( {2a} \right)}^2} - 4 \times 1 \times b} }}{{2 \times 1}}$
On simplifying the value we can write
$x = \dfrac{{ - 2a \pm \sqrt {4{a^2} - 4b} }}{2}$
We will take the common factor out, so it gives :
$x = \dfrac{{ - 2a \pm \sqrt {4({a^2} - b)} }}{2}$
We know that the value
$\sqrt 4 = 2$
So we can write
$x = \dfrac{{ - 2a \pm 2({a^2} - b)}}{2}$
We will simplify the value i.e.
$x = \dfrac{{ - 2a}}{2} \pm \dfrac{{2\sqrt {{a^2} - b} }}{2}$
So it gives the value:
$x = - a \pm \sqrt {{a^2} - b}$

We will separate both the signs i.e. it gives us two values, one is positive and one is negative.
It gives
${x_1} = - a + \sqrt {{a^2} - b} \,,\,{x_2} = - a - \sqrt {{a^2} - b}$
Similarly we will solve for second equation :
${x^2} + 2cx + d = 0$
Again we will use the quadratic formula which is
$\dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}$
Here we have
$b = 2c,c = d,a = 1$
Now by applying the formula we can write
$x = \dfrac{{ - 2c \pm \sqrt {{{\left( {2c} \right)}^2} - 4 \times 1 \times d} }}{{2 \times 1}}$
On simplifying the value we can write
$x = \dfrac{{ - 2c \pm \sqrt {4{c^2} - 4d} }}{2}$
We will take the common factor out, so it gives:
$x = \dfrac{{ - 2c \pm \sqrt {4({c^2} - d)} }}{2}$
We know that the value
$\sqrt 4 = 2$ , so will take this value out .
So we can write
$x = \dfrac{{ - 2a \pm 2({c^2} - d)}}{2}$
We will simplify the value i.e.
$x = \dfrac{{ - 2c}}{2} \pm \dfrac{{2\sqrt {{c^2} - d} }}{2}$
So it gives the value:
$y = - c \pm \sqrt {{c^2} - d}$
Again it gives us two values: :
${y_1} = - c + \sqrt {{c^2} - d} \,,{y_2} = - c - \sqrt {{c^2} - d}$
Now we have the formula:
$(x - {x_1})(x - {x_2})(y - {y_1})(y - {y_2})$ .
By substituting the values in the formula we can write: \left[ {\left\\{ {x - \left( { - a + \sqrt {{a^2} - b} } \right)} \right\\}\left\\{ {x - \left( { - a - \sqrt {{a^2} - b} } \right)} \right\\}} \right] + \left[ {\left\\{ {y - \left( { - c + \sqrt {{c^2} - d} } \right)} \right\\}\left\\{ {y - \left( { - c - \sqrt {{c^2} - d} } \right)} \right\\}} \right]
On simplifying the brackets we can write them as
$\left( {x + a - \sqrt {{a^2} - b} } \right)\left( {x + a + \sqrt {{a^2} - b} } \right) + \left( {y + c - \sqrt {{c^2} - d} } \right)\left( {y + c + \sqrt {{c^2} - d} } \right)$ .
We can see that the first and second part of the expression are of the form $(a + b)(a - b)$.We can apply the formula of $(a + b)(a - b) = {a^2} - {b^2}$. Let us take the first part of the expression i.e.
$\left( {x + a - \sqrt {{a^2} - b} } \right)\left( {x + a + \sqrt {{a^2} - b} } \right)$ .

By comparing with the formula, here we have
$a = x + a,b = \sqrt {{a^2} - b}$
By substituting these in the formula we can write;
\left\\{ {{{\left( {x + a} \right)}^2} - {{\left( {\sqrt {{a^2} - b} } \right)}^2}} \right\\} .
Again we can apply the formula
${(a + b)^2} = {a^2} + {b^2} + 2ab$ .
We will put this value in
${(x + a)^2} = {x^2} + {a^2} + 2ax$
And,
${\left( {\sqrt {{a^2} - b} } \right)^2} = {a^2} - b$
By putting the values together we have:
${x^2} + {a^2} + 2ax - ({a^2} - b)$
It gives us ${x^2} + {a^2} + 2ax - {a^2} + b$ .

Now we have second part,
$\left( {y + c - \sqrt {{c^2} - d} } \right)\left( {y + c + \sqrt {{c^2} - d} } \right)$ .
We will solve this part same as the second part, by applying formula
$(a + b)(a - b) = {a^2} - {b^2}$ .
We should note that in the second part we have
$a = y + c,b = \sqrt {{c^2} - d}$
By substituting these in the formula we can write;
\left\\{ {{{\left( {y + c} \right)}^2} - {{\left( {\sqrt {{c^2} - d} } \right)}^2}} \right\\} .
Again we can apply the formula
${(a + b)^2} = {a^2} + {b^2} + 2ab$ .
We will put this value in
${(y + c)^2} = {y^2} + {c^2} + 2yc$
And, ${\left( {\sqrt {{c^2} - d} } \right)^2} = {c^2} - d$

By putting the values together it gives:
${y^2} + {c^2} + 2yc - ({c^2} - d) = {y^2} + {c^2} + 2yc - {c^2} + d$
Therefore by simplifying this as same as the first part method, it gives us
${y^2} + {c^2} + 2yc - {c^2} + d$ .
Now we will put the values of both bark together in the original expression:
${x^2} + {a^2} + 2ax - {a^2} + b + \left( {{y^2} + {c^2} + 2yc - {c^2} + d} \right)$
On simplifying we have;
${x^2} + {a^2} + 2ax - {a^2} + b + {y^2} + {c^2} + 2yc - {c^2} + d$
It gives us value:
$\therefore {x^2} + {y^2} + 2ax + 2yc + b + d = 0$

Hence the correct option is A.

Note: We should keep in mind that abscissa is the $x -$ coordinate of any point and ordinate is the $y -$ coordinate of any point on the $x - y$ coordinate. We should also know another approach on how to find the roots of the equation. The formula of the sum of the roots is given by the formula $\dfrac{{ - b}}{a}$.Similarly we can calculate the product of the roots $\dfrac{c}{a}$.