If the length of the chord of the circle x^{2 | vedclass

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(B) The equation of the circle is $x^{2}+y^{2}=r^{2}$ with center $O(0,0)$ and radius $r$.Let the chord be $AB$ with length $AB=r$.Let $M$ be the midp

Vedclass · Accepted Answer

$\frac{12}{5}$

Vedclass · Answer

(B) The equation of the circle is $x^{2}+y^{2}=r^{2}$ with center $O(0,0)$ and radius $r$.Let the chord be $AB$ with length $AB=r$.Let $M$ be the midpoint of the chord $AB$. Then $OM \perp AB$.In the right-angled triangle $\Delta OAM$,$OA=r$ (radius) and $AM = \frac{AB}{2} = \frac{r}{2}$.Using the Pythagorean theorem,$OM^{2} = OA^{2} - AM^{2} = r^{2} - (\frac{r}{2})^{2} = r^{2} - \frac{r^{2}}{4} = \frac{3r^{2}}{4}$.Thus,$OM = \frac{r\sqrt{3}}{2}$.The perpendicular distance from the center $(0,0)$ to the line $2x-y+3=0$ is given by $d = \frac{|2(0) - (0) + 3|}{\sqrt{2^{2} + (-1)^{2}}} = \frac{3}{\sqrt{5}}$.Equating $OM = d$,we get $\frac{r\sqrt{3}}{2} = \frac{3}{\sqrt{5}}$.Squaring both sides,$\frac{3r^{2}}{4} = \frac{9}{5}$.Therefore,$r^{2} = \frac{9}{5} 	imes \frac{4}{3} = \frac{12}{5}$.

If the length of the chord of the circle $x^{2}+y^{2}=r^{2}$ $(r>0)$ along the line $y-2x=3$ is $r$,then $r^{2}$ is equal to

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