Question

PQ is a chord of length $8$ cm of a circle of radius $5$ cm. The tangents at P and Q intersect at a point T. Find the length TP.

Answer 1

$\begin{array}{c}letTPisxandTOisy.\\ TO\perp PQ\\ \therefore PO=OQ=4cm\\ Inrightangletrianglepoc\\ \Rightarrow o{c}^2=5^2-4^2\\ \Rightarrow oc=\sqrt{25-16}\\ \Rightarrow oc=\sqrt{9}=3\\ InrigthtriangleCPT\\ \Rightarrow x^2+5^2={\left(y+3\right)}^2\\ \Rightarrow x^2+25=y^2+6y+\mathrm{9.....}\left(1\right)\\ InrigthtrainglePOT\\ \Rightarrow x^2=y^2+4^2\\ \Rightarrow x^2=y^2+\mathrm{16.....}\left(2\right)\\ Substract\left(1\right)from\left(2\right)\\ \Rightarrow 25=6y-7\\ \Rightarrow 25+7=6y\\ \Rightarrow 32=6y\\ \Rightarrow y=\frac{16}{3}cm\\ puty=2cminequation\left(2\right)\\ \Rightarrow x^2={\left(\frac{16}{3}\right)}^2+16\\ \Rightarrow x^2=\frac{256}{9}+16\\ \Rightarrow x^2=16\left(\frac{16+9}{9}\right)\\ \Rightarrow x=\sqrt{16\left(\frac{25}{9}\right)}\\ \Rightarrow x=\frac{20}{3}\end{array}$