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anonymous one year ago 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

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1. anonymous

you got the equation for the line?

2. anonymous

thats all the question give

3. anonymous

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

4. anonymous

|dw:1443226608217:dw|

5. anonymous

??

6. anonymous

yeah good

7. 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

8. anonymous

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

9. anonymous

should get $(25 x^2)/9+(40 x)/3+25 = 9$

10. anonymous

oh yeah, what you got

11. anonymous

multiply both sides by 9

12. anonymous

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

13. anonymous

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

14. anonymous

typo there, i meant $25x^2+120x+144=0$

15. anonymous

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

16. anonymous

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

17. anonymous

@satellite73

18. anonymous

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

19. anonymous

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

20. anonymous

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

21. anonymous

yes it would cut the circle twice

22. anonymous

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

23. anonymous

ok

24. anonymous

thank you

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