a-parabola-y-x-2-15x-36-cuts-the-x-axis-at-P-and-Q-a-circle-is-drawn-through-P-and-Q-so-that-the-origin-is-outside-it-then-find-the-length-of-tangent-to-the-circle-from-0-0-

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Question Number 147009 by gsk2684 last updated on 17/Jul/21

$$aparabolay=x^2-15x+36cutsthe$$$$xaxisatPandQ.acircleisdrawn$$$$throughPandQsothattheorigin$$$$isoutsideit.thenfindthelength$$$$oftangenttothecirclefrom(0,0)?$$

Answered by mr W last updated on 17/Jul/21

$$y=x^2-15x+36=(x-3)(x-12$$$$\Rightarrow P(3,0)$$$$\Rightarrow Q(12,0)$$$$\Rightarrow M(\frac{15}{2},0)$$$$letC(h,0)$$$${OC}^2={\left(\frac{15}{2}\right)}^2+h^2$$$${AC}^2={CP}^2={\left(\frac{12-3}{2}\right)}^2+h^2={\left(\frac{9}{2}\right)}^2+h^2$$$${OA}^2={OC}^2-{AC}^2={\left(\frac{15}{2}\right)}^2-{\left(\frac{9}{2}\right)}^2$$$$\Rightarrow OA=\frac{\sqrt{{15}^2-9^2}}{2}=\frac{12}{2}=6$$$$thatmeansthelengthoftangent$$$$isalways6.$$

Commented by mr W last updated on 17/Jul/21

Commented by mr W last updated on 17/Jul/21

$$method2:$$$$generally$$$${OA}^2=OP\times OQ=3\times 12=36$$$$\Rightarrow OA=6$$$$seeQ137946$$

Commented by gsk2684 last updated on 18/Jul/21

$$thankyou$$

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