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Question Number 137946 by otchereabdullai@gmail.com last updated on 08/Apr/21

From a poin P outside a circle a

tangent is drawn to a circle at T .

from P a line is drawn to circle at

A and B . Find lenth PT giving that

i . PA = 6

AB = 8

ii . PB = 18

PT = 12

Commented by mr W last updated on 08/Apr/21

g e n e r a l l y w e h a v e

P T = \sqrt{P A \times P B}

(t e l l i f y o u n e e d a p r o o f)

Commented by mr W last updated on 08/Apr/21

i .

P T = \sqrt{6 \times (6 + 8)} = 2 \sqrt{21}

i i .

12 = \sqrt{P A \times 18}

\Rightarrow P A = \frac{12^{2}}{18} = 8 \Rightarrow A B = 18 - 8 = 10

Commented by mr W last updated on 08/Apr/21

Commented by otchereabdullai@gmail.com last updated on 08/Apr/21

God bless u profW i will be very glad

to see the proof

Commented by mr W last updated on 08/Apr/21

Commented by mr W last updated on 08/Apr/21

{C D}^{2} = R^{2} - {(\frac{b}{2})}^{2}

{C P}^{2} = {C D}^{2} + {P D}^{2} = R^{2} - {(\frac{b}{2})}^{2} + {(a + \frac{b}{2})}^{2}

{C P}^{2} = {C T}^{2} + {P T}^{2} = R^{2} + t^{2}

R^{2} - {(\frac{b}{2})}^{2} + {(a + \frac{b}{2})}^{2} = R^{2} + t^{2}

- {(\frac{b}{2})}^{2} + {(a + \frac{b}{2})}^{2} = t^{2}

a (a + b) = t^{2}

\Rightarrow t = \sqrt{a (a + b)}

i . e . P T = \sqrt{P A \times P B}

Commented by otchereabdullai@gmail.com last updated on 08/Apr/21

Thanks for the kindness and may the

almigthy Allah add more years to

age