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


Question and Answers Forum

All Questions Topic List
None Questions
Previous in All Question Next in All Question
Previous in None Next in None
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

$Thanks for the kindness and may the$ $almigthy Allah add more years to$ $age$
Terms of Service
Privacy Policy
Contact: info@tinkutara.com

Question and Answers Forum