Chords AB and CD of a circle intersect at a r | vedclass

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(B) Let the point of intersection $P$ be the origin $(0, 0)$. Since the chords intersect at a right angle, we can place $AB$ along the $x$-axis and $C

Vedclass · Accepted Answer

$\frac{\sqrt{65}}{2}$ units

Vedclass · Answer

(B) Let the point of intersection $P$ be the origin $(0, 0)$. Since the chords intersect at a right angle, we can place $AB$ along the $x$-axis and $CD$ along the $y$-axis.Given $AP = 2$, $PB = 6$, $CP = 3$, and $PD = 4$, the coordinates of the endpoints are $A(-2, 0)$, $B(6, 0)$, $C(0, 3)$, and $D(0, -4)$.Let the center of the circle be $O(h, k)$.The perpendicular bisector of $AB$ is $x = \frac{-2 + 6}{2} = 2$.The perpendicular bisector of $CD$ is $y = \frac{3 - 4}{2} = -0.5$.Thus, the center is $O(2, -0.5)$.The radius $r$ is the distance from $O(2, -0.5)$ to $A(-2, 0)$:$r^2 = (2 - (-2))^2 + (-0.5 - 0)^2 = 4^2 + (-0.5)^2 = 16 + 0.25 = 16.25 = \frac{65}{4}$.Therefore, $r = \sqrt{\frac{65}{4}} = \frac{\sqrt{65}}{2}$ units.

Chords $AB$ and $CD$ of a circle intersect at a right angle at point $P$. If the lengths of $AP$, $PB$, $CP$, and $PD$ are $2$, $6$, $3$, and $4$ units respectively, then the radius of the circle is:

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