Let the abscissae of the two points P and Q o | vedclass

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(A) The equation of a circle with diameter endpoints $(x_{1}, y_{1})$ and $(x_{2}, y_{2})$ is given by $(x-x_{1})(x-x_{2}) + (y-y_{1})(y-y_{2}) = 0$.T

Vedclass · Accepted Answer

$12$

Vedclass · Answer

(A) The equation of a circle with diameter endpoints $(x_{1}, y_{1})$ and $(x_{2}, y_{2})$ is given by $(x-x_{1})(x-x_{2}) + (y-y_{1})(y-y_{2}) = 0$.This expands to $x^{2} - (x_{1}+x_{2})x + x_{1}x_{2} + y^{2} - (y_{1}+y_{2})y + y_{1}y_{2} = 0$.Given that $x_{1}, x_{2}$ are roots of $x^{2}-4x-6=0$,we have $x_{1}+x_{2} = 4$ and $x_{1}x_{2} = -6$.Given that $y_{1}, y_{2}$ are roots of $y^{2}+2y-7=0$,we have $y_{1}+y_{2} = -2$ and $y_{1}y_{2} = -7$.Substituting these into the circle equation:$x^{2} - 4x - 6 + y^{2} + 2y - 7 = 0$$x^{2} + y^{2} - 4x + 2y - 13 = 0$.Comparing this with $x^{2} + y^{2} + 2ax + 2by + c = 0$,we get:$2a = -4 \implies a = -2$$2b = 2 \implies b = 1$$c = -13$Therefore,$a+b-c = -2 + 1 - (-13) = -1 + 13 = 12$.

Let the abscissae of the two points $P$ and $Q$ on a circle be the roots of $x^{2}-4x-6=0$ and the ordinates of $P$ and $Q$ be the roots of $y^{2}+2y-7=0$. If $PQ$ is a diameter of the circle $x^{2}+y^{2}+2ax+2by+c=0$,then the value of $(a+b-c)$ is.

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