If the locus of the midpoints of the chords o | vedclass

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(B) The equation of the circle is $x^2+y^2=25$,so the radius $r=5$.Let $C(x_1, y_1)$ be the midpoint of a chord $AB$ that subtends a right angle at th

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$\frac{5}{\sqrt{2}}$

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(B) The equation of the circle is $x^2+y^2=25$,so the radius $r=5$.Let $C(x_1, y_1)$ be the midpoint of a chord $AB$ that subtends a right angle at the origin $O(0, 0)$.In $	riangle OAB$,$OA=OB=5$ and $\angle AOB = 90^\circ$.Since $OC$ is the median to the hypotenuse $AB$ in the right-angled $	riangle OAB$,$OC = \frac{1}{2} AB$.Alternatively,in $	riangle OCB$,$\angle COB = 45^\circ$ and $\angle OCB = 90^\circ$.Thus,$OC = OB \cos(45^\circ) = 5 	imes \frac{1}{\sqrt{2}} = \frac{5}{\sqrt{2}}$.The distance of the midpoint $C(x_1, y_1)$ from the origin is $\sqrt{x_1^2+y_1^2}$.Therefore,$\sqrt{x_1^2+y_1^2} = \frac{5}{\sqrt{2}}$,which implies $x_1^2+y_1^2 = \frac{25}{2}$.Dividing by $\frac{25}{2}$,we get $\frac{x_1^2}{25/2} + \frac{y_1^2}{25/2} = 1$.Comparing this with $\frac{x^2}{a^2} + \frac{y^2}{a^2} = 1$,we have $a^2 = \frac{25}{2}$,so $|a| = \frac{5}{\sqrt{2}}$.

If the locus of the midpoints of the chords of the circle $x^2+y^2=25$,which subtend a right angle at the origin,is given by $\frac{x^2}{a^2}+\frac{y^2}{a^2}=1$,then $|a|=$

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