Prove that the lengths of tangents drawn from an external point to a circle are equal.

Question :Prove that the lengths of tangents drawn from an external point to a circle are equal.

The correct answer :Let O be the center of a circle, and let A and B be two points on the circle. Let P be an external point to the circle, and let PA and PB be the tangents from P to the circle. We want to show that PA = PB.

First, we draw a line segment OP from the external point P to the center O of the circle, and let the point of intersection of OP with the circle be C.

Since OA and OB are radii of the circle, they are equal in length. Also, since OC is a radius of the circle, it is perpendicular to the tangent PA at point A, and to the tangent PB at point B.

Therefore, we have two right triangles, PAC and PBC, where PC is the hypotenuse of both triangles. By the Pythagorean theorem, we have:

PA² = PC² – AC² PB² = PC² – BC²

Subtracting these two equations, we get:

PA² – PB² = (PC² – AC²) – (PC² – BC²)

Simplifying, we get:

PA² – PB² = BC² – AC²

But AC = BC, since they are both radii of the same circle, so AC² = BC². Therefore, we can simplify further:

PA² – PB² = 0

Which means that:

PA² = PB²

Taking the square root of both sides, we get:

PA = PB

Therefore, we have shown that the lengths of the tangents drawn from an external point to a circle are equal.