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III.36 Proposition III.36

If a point be taken outside a circle and from it there fall on the circle two straight lines, and if one of them cut the circle and the other touch it, the rectangle contained by the whole of the straight line which cuts the circle and the straight line intercepted on it outside between the point and the convex circumference will be equal to the square on the tangent.

Proof

Let

P

be the external point,

P T

the tangent at

T

, and

P A B

a secant cutting the circle at

A

(near) and

B

(far); let

O

be the centre and

r

the radius. By III.18,

O T ⊥ P T

. By I.47 in

△ O T P

:

O P^{2} = O T^{2} + P T^{2} = r^{2} + P T^{2}

, hence

P T^{2} = O P^{2} - r^{2}

. Let

M

be the midpoint of

A B

; by III.3,

O M ⊥ A B

. Apply II.6 to

A B

bisected at

M

and produced to

P

(so

P

is on line

A B

extended beyond the near intersection

A

):

P A \cdot P B + M A^{2} = M P^{2}

. By I.47 in

△ O M A

,

M A^{2} = r^{2} - O M^{2}

; in

△ O M P

,

M P^{2} = O P^{2} - O M^{2}

. Subtracting:

P A \cdot P B = O P^{2} - r^{2}

. Comparing:

P T^{2} = O P^{2} - r^{2} = P A \cdot P B

.

Figure

$Proposition III.36. From an external point $P$, the tangent $PT$ and any secant $PAB$ satisfy $PT^2 = PA \cdot PB$ (the power of $P$ with respect to the circle).$

Proposition III.36. From an external point

P

, the tangent

P T

and any secant

P A B

satisfy

P T^{2} = P A \cdot P B

(the power of

lines 74–74 in main.tex

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III.36 Proposition III.36

If a point be taken outside a circle and from it there fall on the circle two straight lines, and if one of them cut the circle and the other touch it, the rectangle contained by the whole of the straight line which cuts the circle and the straight line intercepted on it outside between the point and the convex circumference will be equal to the square on the tangent.

Proof

Let

P

be the external point,

P T

the tangent at

T

, and

P A B

a secant cutting the circle at

A

(near) and

B

(far); let

O

be the centre and

r

the radius. By III.18,

O T ⊥ P T

. By I.47 in

△ O T P

:

O P^{2} = O T^{2} + P T^{2} = r^{2} + P T^{2}

, hence

P T^{2} = O P^{2} - r^{2}

. Let

M

be the midpoint of

A B

; by III.3,

O M ⊥ A B

. Apply II.6 to

A B

bisected at

M

and produced to

P

(so

P

is on line

A B

extended beyond the near intersection

A

):

P A \cdot P B + M A^{2} = M P^{2}

. By I.47 in

△ O M A

,

M A^{2} = r^{2} - O M^{2}

; in

△ O M P

,

M P^{2} = O P^{2} - O M^{2}

. Subtracting:

P A \cdot P B = O P^{2} - r^{2}

. Comparing:

P T^{2} = O P^{2} - r^{2} = P A \cdot P B

.

Figure

Proposition III.36. From an external point $P$, the tangent
$PT$ and any secant $PAB$ satisfy $PT^2 = PA \cdot PB$ (the power of
$P$ with respect to the circle). — Proposition III.36. From an external point $P$ , the tangent $P T$ and any secant $P A B$ satisfy $P T^{2} = P A \cdot P B$ (the power of

lines 74–74 in main.tex

P

with respect to the circle).