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Question: The abscissa of two points A and B are the roots of equation \[{x^2} + 2ax - {b^2} = 0\] and their o...

The abscissa of two points A and B are the roots of equation x2+2axb2=0{x^2} + 2ax - {b^2} = 0 and their ordinates are the roots of the equation x2+2pxq2=0{x^2} + 2px - {q^2} = 0. The equation of the circle with AB as diameter is
A.x2+y2+2ax+2py+b2+q2=0{x^2} + {y^2} + 2ax + 2py + {b^2} + {q^2} = 0
B.x2+y22ax2pyb2q2=0{x^2} + {y^2} - 2ax - 2py - {b^2} - {q^2} = 0
C.x2+y2+2ax+2pyb2q2=0{x^2} + {y^2} + 2ax + 2py - {b^2} - {q^2} = 0
D.None of these

Explanation

Solution

The equations given here are of the form of quadratic equations. We will first find the value of their roots with the help of relation between the roots of the equation because that is going to give us the value of endpoints or coordinates of the two points A and B on the circle and are the endpoints of diameter AB. Then we can find the equation of the circle with the help of a diametric equation.

Formula used:
Suppose there are two points on a circle (x1,y1)\left( {{x_1},{y_1}} \right) and (x2,y2)\left( {{x_2},{y_2}} \right) such that they lie on the opposite ends of the same diameter, then the equation of the circle can be written as
(xx1)(xx2)+(yy1)(yy2)=0.(x - {x_1})(x - {x_2}) + (y - {y_1})(y - {y_2}) = 0.

Complete step-by-step answer:
Let us first find the roots of both the equations.

Equationx2+2axb2=0{x^2} + 2ax - {b^2} = 0x2+2pxq2=0{x^2} + 2px - {q^2} = 0
RootsLet the roots be x1,x2{x_1},{x_2}Let the roots be y1,y2{y_1},{y_2}
Comparing with quadratic equation a=1, b=2a and c=b2 - {b^2}Comparing with quadratic equationa=1, b=2p and c= q2 - {q^2}
Relation between the roots{x_1}.{x_2} = \dfrac{c}{a} = - {b^2}$$$${x_1} + {x_2} = \dfrac{{ - b}}{a} = - 2a{y_1}.{y_2} = \dfrac{c}{a} = - {q^2}$$$${y_1} + {y_2} = \dfrac{{ - b}}{a} = - 2p

Now let’s find the equation of the circle with the help of diametric form,
(xx1)(xx2)+(yy1)(yy2)=0.(x - {x_1})(x - {x_2}) + (y - {y_1})(y - {y_2}) = 0.
On multiplying the brackets,
x2xx2xx1+x1x2+y2yy2yy1+y1y2=0\Rightarrow {x^2} - x{x_2} - x{x_1} + {x_1}{x_2} + {y^2} - y{y_2} - y{y_1} + {y_1}{y_2} = 0
Rearranging the terms,
x2+y2x(x1+x2)y(y1+y2)+x1x2+y1y2=0\Rightarrow {x^2} + {y^2} - x\left( {{x_1} + {x_2}} \right) - y\left( {{y_1} + {y_2}} \right) + {x_1}{x_2} + {y_1}{y_2} = 0
Replacing the terms with those of relation between the roots,
x2+y2+2ax+2pyb2q2=0\Rightarrow {x^2} + {y^2} + 2ax + 2py - {b^2} - {q^2} = 0
This is the equation of the circle.
Thus option C is the correct option.
Note: Here note that abscissa and ordinates are the roots of the equation. But since we are using diametric form we do not need the coordinates. But if we use any other form like where we need coordinates of the points then only need to find the roots.