The radical axis of the co-axial system of ci | vedclass

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Question

(D) The radical axis of a co-axial system of circles with limiting points $A(1, 2)$ and $B(-2, 1)$ is the perpendicular bisector of the line segment $

Vedclass · Accepted Answer

$3x + y = 0$

Vedclass · Answer

(D) The radical axis of a co-axial system of circles with limiting points $A(1, 2)$ and $B(-2, 1)$ is the perpendicular bisector of the line segment $AB$.First,find the slope of the line segment $AB$:$m_{AB} = \frac{1 - 2}{-2 - 1} = \frac{-1}{-3} = \frac{1}{3}$.The slope of the perpendicular bisector is $m_{\perp} = -\frac{1}{m_{AB}} = -3$.The midpoint of $AB$ is $M = \left(\frac{1 - 2}{2}, \frac{2 + 1}{2}\right) = \left(-\frac{1}{2}, \frac{3}{2}\right)$.The equation of the radical axis is $y - y_1 = m_{\perp}(x - x_1)$:$y - \frac{3}{2} = -3(x + \frac{1}{2})$$y - \frac{3}{2} = -3x - \frac{3}{2}$$3x + y = 0$.Thus,option $D$ is correct.

The radical axis of the co-axial system of circles with limiting points $(1, 2)$ and $(-2, 1)$ is

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