The radius of the circle having three chords | vedclass

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(C) Let the circle be $(x-h)^2 + (y-k)^2 = r^2$. The distance from the center $(h, k)$ to the chords must be equal to the radius $r$ if the chords wer

Vedclass · Accepted Answer

$\frac{5}{\sqrt{13}}$

Vedclass · Answer

(C) Let the circle be $(x-h)^2 + (y-k)^2 = r^2$. The distance from the center $(h, k)$ to the chords must be equal to the radius $r$ if the chords were tangents,but here they are chords. However,the problem implies these lines are chords of equal length or specific geometry. Assuming the lines are chords of equal length $2L$,the distance from the center to each line is $d = \sqrt{r^2 - L^2}$. For the lines $x=0$,$x-y=0$,and $2x+3y-10=0$,the distances from $(h, k)$ are $d_1 = |h|$,$d_2 = \frac{|h-k|}{\sqrt{2}}$,and $d_3 = \frac{|2h+3k-10|}{\sqrt{13}}$. Setting $d_1 = d_2 = d_3 = d$,we solve for $h$ and $k$. $|h| = \frac{|h-k|}{\sqrt{2}} \implies h^2 = \frac{h^2-2hk+k^2}{2} \implies h^2+2hk-k^2=0$. Solving this system with the third distance leads to the radius $r = \frac{5}{\sqrt{13}}$.

The radius of the circle having three chords along the $y$-axis,the line $y=x$,and the line $2x+3y=10$ is

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