The x-coordinates of two points A and B are t | vedclass

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(A) Let $A = (x_{1}, y_{1})$ and $B = (x_{2}, y_{2})$.From the given equations,we have:$x_{1} + x_{2} = -2a$ and $x_{1}x_{2} = -b^{2}$$y_{1} + y_{2} =

Vedclass · Accepted Answer

$\sqrt{a^{2} + b^{2} + p^{2} + q^{2}}$

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(A) Let $A = (x_{1}, y_{1})$ and $B = (x_{2}, y_{2})$.From the given equations,we have:$x_{1} + x_{2} = -2a$ and $x_{1}x_{2} = -b^{2}$$y_{1} + y_{2} = -2p$ and $y_{1}y_{2} = -q^{2}$The equation of the circle with $AB$ as diameter is $(x - x_{1})(x - x_{2}) + (y - y_{1})(y - y_{2}) = 0$.Expanding this,we get $x^{2} - (x_{1} + x_{2})x + x_{1}x_{2} + y^{2} - (y_{1} + y_{2})y + y_{1}y_{2} = 0$.Substituting the values,we get $x^{2} + 2ax - b^{2} + y^{2} + 2py - q^{2} = 0$,which is $x^{2} + y^{2} + 2ax + 2py - (b^{2} + q^{2}) = 0$.The standard form of a circle is $x^{2} + y^{2} + 2gx + 2fy + c = 0$,where the radius is $\sqrt{g^{2} + f^{2} - c}$.Here,$g = a$,$f = p$,and $c = -(b^{2} + q^{2})$.Therefore,the radius is $\sqrt{a^{2} + p^{2} - (-(b^{2} + q^{2}))} = \sqrt{a^{2} + p^{2} + b^{2} + q^{2}}$.

The $x$-coordinates of two points $A$ and $B$ are the roots of the equation $x^{2} + 2ax - b^{2} = 0$,and their $y$-coordinates are the roots of the equation $y^{2} + 2py - q^{2} = 0$. Find the radius of the circle with $AB$ as its diameter.

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