Given a circle with center O, radius $r$, and a point $P$ outside the circle such that $OP = d$, find the length of the tangent segment from $P$ to the circle.

Let the tangent point be T. In the right triangle OTP, OT is the radius (r), and OP is the distance (d). The length of the tangent segment PT can be found using the Pythagorean theorem:

$OP^2 = PT^2 + OT^2$

Substitute the known values:

$d^2 = PT^2 + r^2$

Solve for PT:

$PT^2 = d^2 – r^2$

$PT = \sqrt{d^2 – r^2}$

Therefore, the length of the tangent segment from P to the circle is:

$\boxed{\sqrt{d^2 – r^2}}$

Consider the radius OT and the tangent PT forming a right triangle with OP:

$OP^2 = OT^2 + PT^2$

Given that OT = r and OP = d, we substitute these values:

$d^2 = r^2 + PT^2$

Rearrange to solve for PT:

$PT^2 = d^2 – r^2$

$PT = sqrt{d^2 – r^2}$

The length of the tangent segment is:

$oxed{sqrt{d^2 – r^2}}$

Using the Pythagorean theorem:

$d^2 = r^2 + PT^2$

So,

$PT = sqrt{d^2 – r^2}$

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