In the following figure PR is a secant and PT is a tangent to the circle If $P T = 6 c m$ and $P T = 6 c m$ and $Q R = 5 c m$ then PQ = ___cm

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Solution

PQR is a secant, PT is a tangent
$∴ P Q \times P R = P T^{2} . . . . . (1)$
Given, $P T = 6 Q R = 5$
Let $P Q = x$
$∴ P R = P Q + Q R = x + 5$
from (1)
$P Q \times P R = P T^{2}$
$x (x + 5) = 6^{2}$
$x^{2} + 5 x = 36$
$x^{2} + 5 x - 36 = 0$
$(x + 9) (x - 4) = 0$
$x + 9 = 0; x - 4 = 0$
$x = - 9; x = 4$
Since length is always positive
$∴ x = P Q = 4 c m$

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