Let S be the circumcircle of the triangle for | vedclass

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(A) The line $x-2y-4=0$ intersects the coordinate axes at $A(4, 0)$ and $B(0, -2)$.Since the triangle is formed with the coordinate axes,it is a right

Vedclass · Accepted Answer

$5-\sqrt{5}$

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(A) The line $x-2y-4=0$ intersects the coordinate axes at $A(4, 0)$ and $B(0, -2)$.Since the triangle is formed with the coordinate axes,it is a right-angled triangle with the right angle at the origin $O(0, 0)$.The hypotenuse is the segment $AB$,so the center of the circumcircle $S$ is the midpoint of $AB$.Center $C = \left(\frac{4+0}{2}, \frac{0-2}{2}\right) = (2, -1)$.The radius $r$ is the distance from $C(2, -1)$ to $O(0, 0)$,so $r = \sqrt{(2-0)^2 + (-1-0)^2} = \sqrt{4+1} = \sqrt{5}$.The distance from $P(-2, -4)$ to the center $C(2, -1)$ is $PC = \sqrt{(2 - (-2))^2 + (-1 - (-4))^2} = \sqrt{4^2 + 3^2} = \sqrt{16+9} = 5$.The minimum distance from a point $P$ outside the circle to a point $Q$ on the circle is $PQ = PC - r$.Thus,$PQ = 5 - \sqrt{5}$.

Let $S$ be the circumcircle of the triangle formed by the line $x-2y-4=0$ with the coordinate axes. If $P(-2, -4)$ is a point in the plane of the circle $S$ and $Q$ is a point on $S$ such that the distance between $P$ and $Q$ is the least,then $PQ=$

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