By finding the points of intersection, show that the line through (0,5) and (3,9) is tangent to the circle x*2+y^2=9

- anonymous

- chestercat

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- anonymous

you got the equation for the line?

- anonymous

thats all the question give

- anonymous

ok let me rephrase the question
can you find the equation for the line?

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## More answers

- anonymous

|dw:1443226608217:dw|

- anonymous

??

- anonymous

yeah good

- anonymous

so now to find the point of intersection replace \(y\) in \(x^2+y^2=9\) by \(\frac{4}{3}x+5\) and solve
\[x^2+(\frac{4}{3}x+5)^2=9\] and see that there is only one solution to that quadratic

- anonymous

x^2+(16/9)x^2+(40/3)x+25=9
(25/9)x^2 + (40/3)x=-16

- anonymous

should get \[(25 x^2)/9+(40 x)/3+25 = 9\]

- anonymous

oh yeah, what you got

- anonymous

multiply both sides by 9

- anonymous

\[25x^2+120x+144x=0\] which , by some miracle , is a perfect square

- anonymous

\[(5x+12)^2=0\\
x=-\frac{12}{5}\] is the only solution

- anonymous

typo there, i meant \[25x^2+120x+144=0\]

- anonymous

ok so thats how we show that the line through the given points is tangent to the circle?

- anonymous

x^2+y^2=9
-12/5+y^2=9
y^2=57/5
y=sq.rt (285)/5

- anonymous

- anonymous

this\[25x^2+120x+144=0\]s the quadratic equation you get

- anonymous

the fact that it has one solution (not 2, not 0) is what tells you that the line is tangent to the graph

- anonymous

ok and if it had 2 solutions then its not tangent to the graph?

- anonymous

yes it would cut the circle twice

- anonymous

and if you had no solutions, it would not cross the circle at all

- anonymous

ok

- anonymous

thank you

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