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After using the property we’ll get an equation and solving that equation we’ll get the value of the radius of the circle.

Given data: \[Length{\text{ }}of{\text{ }}tangent = 8cm\]

The distance of the point from centre=17cm

Let the centre of the circle be ‘O’ and the point from which the tangent is made be ‘R’, the point tangent touches the circle be ‘P’.

We know that tangents are always perpendicular to the radius made from that point.

Therefore, \[OP \bot PR\] or $\angle P = {90^ \circ }$

Using Pythagoras theorem

i.e., in a right-angled triangle ABC with right angle at A, $B{C^2} = A{B^2} + C{A^2}$

Therefore, Using Pythagoras theorem in triangle OPR where $\angle P = {90^ \circ }$

$ \Rightarrow O{R^2} = P{R^2} + P{O^2}$

Substituting the value of OR and PR

$ \Rightarrow {17^2} = {8^2} + P{O^2}$

On Separating the unknown term,

$ \Rightarrow P{O^2} = {17^2} - {8^2}$

On squaring we get,

$ \Rightarrow P{O^2} = 289 - 64$

$ \Rightarrow P{O^2} = 225$

Taking positive square root on both sides, we get,

$\therefore PO = 15cm$

1.A tangent of a circle always touches the circle at a single point.

2.Tangent is always perpendicular to the radius made at the point of tangency.

3.The length of two tangents drawn to a single point to a circle is always equal.