If (4,7) and (-2,-1) are ends of a diameter o | vedclass

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(D) The end points of the diameter are $(4,7)$ and $(-2,-1)$.The equation of the circle is given by $(x-x_1)(x-x_2) + (y-y_1)(y-y_2) = 0$.Substituting

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$8$

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(D) The end points of the diameter are $(4,7)$ and $(-2,-1)$.The equation of the circle is given by $(x-x_1)(x-x_2) + (y-y_1)(y-y_2) = 0$.Substituting the values,we get $(x-4)(x+2) + (y-7)(y+1) = 0$.Expanding this,we have $x^2 - 2x - 8 + y^2 - 6y - 7 = 0$,which simplifies to $x^2 + y^2 - 2x - 6y - 15 = 0$.The general equation of a circle is $x^2 + y^2 + 2gx + 2fy + c = 0$.Comparing the equations,we get $g = -1$,$f = -3$,and $c = -15$.The length of the $X$-intercept is given by the formula $2\sqrt{g^2 - c}$.Substituting the values,$AB = 2\sqrt{(-1)^2 - (-15)} = 2\sqrt{1 + 15} = 2\sqrt{16} = 2 	imes 4 = 8$.Thus,$AB = 8$.

If $(4,7)$ and $(-2,-1)$ are ends of a diameter of a circle which intersects $X$-axis at $A$ and $B$,then $AB$ is equal to

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