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Two tangents $\overline{PA}$ and $\overline{PB}$ are drawn to a circle, where $P$ lies outside the circle, and $A$ and $B$ lie on the circle.  The length of $\overline{AB}$ is $4,$ and the circle has a radius of $5.$  Find the length $AB.$

 Apr 22, 2024
 #1
avatar+14943 
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\(The\ length\ of\ \overline{AB}\ is\ 4,\ find\ the\ length\ AB.\\ What's\ this\ nonsense?\)

 

laugh !

.
 Apr 22, 2024
edited by asinus  Apr 22, 2024
 #2
avatar+14943 
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The length of AP is 4, and the circle has a radius of 5.  Find the length AB.

 

\(f_{circle}(x)=\sqrt{5^2-x^2}\\ \color{blue}A\ (0,5)\\ \color{blue}P\ (4,5)\\ f_{PB}(x)=-\sqrt{4^2-(x-4)^2}+5=-\sqrt{16-x^2+8x-16}+5\\ f_{PB}(x)=-\sqrt{8x-x^2}+5 \)

 

\(\sqrt{25-x^2}=-\sqrt{8x-x^2}+5\\ \sqrt{5-x}\sqrt{x+5}=5-\sqrt{8-x}\sqrt{x}\\ (WolframAlpha)\\ x_{B}\in \{0,\frac{200}{41}\}\\ y_b=\sqrt{25-4.\overline{87804}\ ^2}=1.0976\\ B\ (\frac{200}{41},\sqrt{20.\overline{12195}}) \\ \color{blue}B\ (4.\overline{87804},1.0976)\)

 

 \(\overline{AB}= \sqrt{(y_A-y_B)^2+(x_B-x_A)^2}=\sqrt{(5-1.0976)^2+(4.\overline{87804}-0)^2}\\ \color{blue}\overline{AB}=6.2469\)

 

\(\color{blue}The\ length\ \overline{AB}\ is\ 6.2469.\)  

 

laugh !

 Apr 22, 2024
edited by asinus  Apr 22, 2024
edited by asinus  Apr 22, 2024
edited by asinus  Apr 22, 2024

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