Let P and Q be two external points of the cir | vedclass

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(C) Let the coordinates of points $P$ and $Q$ be $(h, k)$ and $(p, q)$ respectively.The equation of the chord of contact of the point $P(h, k)$ with r

Vedclass · Accepted Answer

$\sqrt{l_1^2+l_2^2}$

Vedclass · Answer

(C) Let the coordinates of points $P$ and $Q$ be $(h, k)$ and $(p, q)$ respectively.The equation of the chord of contact of the point $P(h, k)$ with respect to the circle $x^2+y^2=a^2$ is $xh+yk=a^2$.Since this chord passes through $Q(p, q)$,we have $ph+qk=a^2$.The lengths of the tangents from $P$ and $Q$ are given by $l_1 = \sqrt{h^2+k^2-a^2}$ and $l_2 = \sqrt{p^2+q^2-a^2}$.Squaring these,we get $l_1^2 = h^2+k^2-a^2$ and $l_2^2 = p^2+q^2-a^2$.The distance $PQ$ is given by $PQ = \sqrt{(h-p)^2+(k-q)^2} = \sqrt{h^2+k^2+p^2+q^2-2(hp+kq)}$.Substituting $hp+kq=a^2$,we get $PQ = \sqrt{(h^2+k^2)+(p^2+q^2)-2a^2}$.Substituting $h^2+k^2 = l_1^2+a^2$ and $p^2+q^2 = l_2^2+a^2$,we get $PQ = \sqrt{(l_1^2+a^2)+(l_2^2+a^2)-2a^2} = \sqrt{l_1^2+l_2^2}$.Thus,option $(C)$ is correct.

Let $P$ and $Q$ be two external points of the circle $S \equiv x^2+y^2-a^2=0$. Let the chord of contact of the point $P$ with respect to the circle $S=0$ pass through $Q$. If $l_1$ and $l_2$ are the lengths of the tangents drawn from $P$ and $Q$ to the circle $S=0$,then $PQ=$

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