Question

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Answer 1

Answer: 4

Step-by-step explanation:

Given

PA=6, PD=4, BE=5

We know, If PEB is a secant that intersects the circle at E and B and PA be a tangent at A, then

$PA^2=PE\cdot PB$

Substitute the given values

$\Rightarrow 6^2=(PE+BE)\cdot(PE)\\\Rightarrow 36=(PE+5)\cdot(PE)$

Take $PE=x$

$\Rightarrow 36=x^2+5x\\\Rightarrow x^2+5x-36=0\\\Rightarrow x^2+9x-5x-36=0\\\Rightarrow (x+9)(x-4)=0\\\Rightarrow x=-9,4$

Discard $x=-9$

$\therefore PE=4$