The common chord of two intersecting circles | vedclass

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(C) Let the common chord be $AB$ and let it intersect the line joining the centres $O_1$ and $O_2$ at $M$. Let $PM = h$ be half the length of the comm

Vedclass · Accepted Answer

$\sqrt{2}$ and $2$

Vedclass · Answer

(C) Let the common chord be $AB$ and let it intersect the line joining the centres $O_1$ and $O_2$ at $M$. Let $PM = h$ be half the length of the common chord.In $	riangle O_1PM$,$\angle PO_1M = 90^\circ / 2 = 45^\circ$. Thus,$h = O_1M 	an 45^\circ = O_1M$.In $	riangle O_2PM$,$\angle PO_2M = 60^\circ / 2 = 30^\circ$. Thus,$h = O_2M 	an 30^\circ = O_2M / \sqrt{3}$,which implies $O_2M = h\sqrt{3}$.The distance between the centres is $O_1M + O_2M = h + h\sqrt{3} = h(1 + \sqrt{3})$.Given $O_1O_2 = \sqrt{3} + 1$,we have $h(1 + \sqrt{3}) = \sqrt{3} + 1$,so $h = 1$.The radius $r_1$ of $c_1$ is $O_1P = h / \sin 45^\circ = 1 / (1/\sqrt{2}) = \sqrt{2}$.The radius $r_2$ of $c_2$ is $O_2P = h / \sin 30^\circ = 1 / (1/2) = 2$.Therefore,the radii are $\sqrt{2}$ and $2$.

The common chord of two intersecting circles $c_1$ and $c_2$ subtends angles of $90^\circ$ and $60^\circ$ at their respective centres. If the distance between their centres is $\sqrt{3} + 1$,then the radii of $c_1$ and $c_2$ are:

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