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 $AB$. Thus,$OP$ is perpendicular to $AB$.The slope of $OP$ is $m_{OP} = \frac{6-0}{-5-0} = -\frac{6}{5}$.Since $AB \perp OP$,the slope of line $l$ (which is $AB$) is $m_l = -\frac{1}{m_{OP}} = \frac{5}{6}$.The equation of line $l$ is $y - y_1 = m(x - x_1)$. However,we can use the form $5x - 6y + k = 0$.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 $AB$),$OA^2 = OM^2 + AM^2$. Here $OA = \sqrt{61}$ and $AM = \sqrt{PA^2 - PM^2}$.First,find $PM$. $P$ is $(-5, 6)$,$O$ is $(0,0)$,so $OP = \sqrt{(-5)^2+6^2} = \sqrt{61}$.Since $PA=10$ and $AM^2 = PA^2 - PM^2$,we need $PM$. $M$ is the projection of $O$ on $l$. $OM = d = \frac{|k|}{\sqrt{61}}$.In $	riangle OMP$,$\angle OMP = 90^\circ$,so $PM^2 = OP^2 - OM^2 = 61 - d^2$.$AM^2 = PA^2 - PM^2 = 100 - (61 - d^2) = 39 + d^2$.Also $AM^2 = OA^2 - OM^2 = 61 - d^2$.Equating: $39 + d^2 = 61 - d^2$ $\Rightarrow 2d^2 = 22$ $\Rightarrow d^2 = 11$.$d = \frac{|k|}{\sqrt{61}} = \sqrt{11} \Rightarrow |k| = \sqrt{61 	imes 11} = \sqrt{671}$. This does not match options.Re-evaluating: The line $l$ passes through $A$ and $B$. The power of point $P$ is $PA \cdot PB = 100$. Also $P$ is $(-5,6)$,$x^2+y^2-61=0$. Power is $(-5)^2+6^2-61 = 25+36-61 = 0$. This means $P$ is on the circle. If $P$ is on the circle,$PA=PB=10$ implies $AB$ is a chord. The line $l$ is the chord of contact if $P$ were outside,but here $P$ is on the circle. The line $l$ must be $5x-6y+k=0$. Testing $5x-6y+11=0$,distance from $(0,0)$ is $11/\sqrt{61}$. $AM^2 = 61 - 121/61 = (3721-121)/61 = 3600/61$. $AM = 60/\sqrt{61}$. $PA^2 = PM^2 + AM^2$. $PM$ is distance from $(-5,6)$ to line $5x-6y+11=0$,$PM = |5(-5)-6(6)+11|/\sqrt{61} = |-25-36+11|/\sqrt{61} = 50/\sqrt{61}$. $PA^2 = 2500/61 + 3600/61 = 6100/61 = 100$. Thus $PA=10$. Correct option is $(c)$.

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

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