A line l meets the circle x^2+y^2=61 at point | vedclass

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(C) The circle is $x^2+y^2=61$ with center $O(0,0)$ and radius $r=\sqrt{61}$.Since $PA=PB=10$,$P$ lies on the perpendicular bisector of chord $AB$. Th

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$5x-6y+11=0$

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(C) The circle is $x^2+y^2=61$ with center $O(0,0)$ and radius $r=\sqrt{61}$.Since $PA=PB=10$,$P$ lies on the perpendicular bisector of chord $AB$. The center $O$ also lies on the perpendicular bisector of chord $AB$. Thus,the line $OP$ is perpendicular to the line $l$.The slope of $OP$ is $m_{OP} = \frac{6-0}{-5-0} = -\frac{6}{5}$.Since $l \perp OP$,the slope of line $l$ is $m_l = -\frac{1}{m_{OP}} = \frac{5}{6}$.Let the equation of line $l$ be $5x-6y+k=0$.The distance from the center $O(0,0)$ to the chord $AB$ is $d = \frac{|k|}{\sqrt{5^2+(-6)^2}} = \frac{|k|}{\sqrt{61}}$.In $	riangle OMA$ (where $M$ is the midpoint of $AB$),$OA^2 = OM^2 + AM^2$.$AM^2 = PA^2 - PM^2$. Since $P(-5,6)$,$OP = \sqrt{(-5)^2+6^2} = \sqrt{61}$.Let $M$ be the projection of $O$ on $l$. $OM = d = \frac{|k|}{\sqrt{61}}$.$AM^2 = r^2 - OM^2 = 61 - \frac{k^2}{61}$.Also $PM^2 = OP^2 - OM^2 = 61 - \frac{k^2}{61}$.Given $PA=10$,$AM^2 = PA^2 - PM^2 = 100 - (61 - \frac{k^2}{61}) = 39 + \frac{k^2}{61}$.Equating $AM^2$: $61 - \frac{k^2}{61} = 39 + \frac{k^2}{61} \implies 22 = \frac{2k^2}{61} \implies k^2 = 11 	imes 61 = 671$.Re-evaluating: The line $l$ passes through $M$. $M$ is the projection of $O$ on $l$. $PM = \sqrt{PA^2 - AM^2} = \sqrt{100 - (61 - d^2)} = \sqrt{39+d^2}$.Since $P, M, O$ are collinear,$PM = |PO - OM| = |\sqrt{61} - d|$.$39+d^2 = 61 - 2d\sqrt{61} + d^2 \implies 2d\sqrt{61} = 22 \implies d = \frac{11}{\sqrt{61}}$.Since $d = \frac{|k|}{\sqrt{61}}$,$|k|=11$. Thus $k = \pm 11$. Checking the options,$5x-6y+11=0$ matches.

$A$ line $l$ meets the circle $x^2+y^2=61$ at points $A$ and $B$. If $P(-5, 6)$ is a point such that $PA=PB=10$,then the equation of line $l$ is:

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