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The abscissae of the two points A and B are the roots of the equationx^2+4 a x-2 b^2=0 and their ordinates are the roots of the equation 2 y^2+2 p y-q^2=0Which of the following is the equation of the circle with AB as the diameter ?

 

Option: 1

x^2+y^2+2 a x+p y+\left(b^2-q^2\right) / 2=0


Option: 2

x^2+y^2+2 a x+p y+\left(b^2+q^2\right) / 2=0


Option: 3

x^2+y^2-2 a x-p y+\left(b^2-q^2\right) / 2=0


Option: 4

x^2+y^2+2 a x+p y-\left(b^2-q^2\right) / 2=0


Answers (1)

best_answer

in the equation    x^2+4 a x-2 b^2=0 using the quadratic formula:

\begin{aligned} & x=\left(-4 a \pm \sqrt{\left.\left(16 a^2+8 b^2\right)\right) / 2}\right. \\ & \left.x=-2 a \pm \sqrt{(} 4 a^2+2 b^2\right) \\ & \end{aligned}

So the abscissae of A and B are:


\begin{gathered} x_A=-2 a+\sqrt{(}\left(4 a^2+2 b^2\right) \\ \left.x_B=-2 a-\sqrt{(} 4 a^2+2 b^2\right) \end{gathered}

In the equation2 y^2+2 p y-q^2=0

\left.y=\left(-p \pm \sqrt{(} p^2+2 q^2\right)\right) / 2

So the ordinates of A and B are:

\begin{gathered} \left.y_A=\left(-p+\sqrt{(} p^2+2 q^2\right)\right) / 2 \\ \left.y_B=\left(-p-\sqrt{(} p^2+2 q^2\right)\right) / 2 \end{gathered}

The midpoint of AB is the center of the circle, which is:


\begin{aligned} & C=\left(\left(x_A+x_B\right) / 2,\left(y_A+y_B\right) / 2\right) \\ = & (-2 a,-p / 2) \end{aligned}

The radius of the circle is half the length of AB, which is:


\begin{aligned} & r=A B / 2=\sqrt{\left(x_B-x_A\right)^2+\left(y_B-y_A\right)^2} / 2 \\ & \left.\left.\left.=\sqrt{(} 4 a \sqrt{(} 4 a^2+2 b^2\right)\right)^2+\left(2 p \sqrt{(} p^2+2 q^2\right)\right)^2 /(4 * 2) \\ & =\sqrt{\left(16 a^2\left(4 a^2+2 b^2\right)+4 p^2\left(p^2+2 q^2\right)\right) / 8} \\ & =\sqrt{\left(16 a^4+8 a^2 b^2+4 p^4+8 p^2 q^2\right) / 8} \\ & =\sqrt{4}\left(a^2+p^2\right)^2+8 a^2 b^2-4 p^2\left(a^2+2 q^2\right) / 8 \\ & \end{aligned}

Therefore, the equation of the circle is:


x^2+y^2+2 a x+p y+\left(b^2-q^2\right) / 2=0

 

Posted by

Nehul

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