Question

PQ is a chord of length 4.8 cm of a circle of radius 3 cm. The tangent at P and Q intersect at a point T as shown in the figure. Find the length of TP.

Answer 1

Given: Radius of the circle, $OP=3cm$ and Chord, $PQ=4.8cm$
Let $TR=y$ and $TP=x$
We know that the perpendicular drawn from the center to the chord bisects it.
$\therefore PR=RQ$
Now, $PR+RQ=4.8$
$\Rightarrow PR+PR=4.8$
$\Rightarrow PR=2.4cm$
Now, in right triangle POR,
By Using Pythagoras theorem, we have
$P{O}^2=O{R}^2+P{R}^2$
$\Rightarrow 3^2=O{R}^2+(2.4{)}^2$
$\Rightarrow O{R}^2=3.24$
$\Rightarrow OR=1.8cm$
Now, in right triangle TPR,
By Using Pythagoras theorem, we have
$T{P}^2=T{R}^2+P{R}^2$
$\Rightarrow x^2=y^2+(2.4{)}^2$
$\Rightarrow x^2=y^2+\mathrm{5.76.......}\left(1\right)$
Again, in right triangle TPO
By Using Pythagoras theorem, we have
$T{O}^2=T{P}^2+P{O}^2$
$\Rightarrow (y+1.8{)}^2=x^2+3^2$
$\Rightarrow y^2+3.6y+3.24=x^2+9$
$\Rightarrow y^2+3.6y=x^2+\mathrm{5.76.......}\left(2\right)$
On Solving eq. (1) and (2), we get
$x=4cm$ and $y=3.2cm$
$\therefore TP=4cm$