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anonymous

  • one year ago

The question goes: A circle, Centre of the origin, radius 5, has equation: x^2+y^2=25 Find the equation of the tangent to the circle that passes through the point(3,4) How do I solve this?

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  1. MrNood
    • one year ago
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    first of all Can you see whether the point given is ON the circle? If so can you work out the slope of the line from th ecentre to the point?....

  2. anonymous
    • one year ago
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    yeah the point given is on the circle, but from here Im not sure where to start or go.

  3. MrNood
    • one year ago
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    so what is the slope of the radius from the centre to that point? |dw:1434550103103:dw|

  4. anonymous
    • one year ago
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    is it 5?

  5. MrNood
    • one year ago
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    no - that is the radius .|dw:1434550282792:dw| slope is y2-y1/x2-x1 centre is (0,0)

  6. anonymous
    • one year ago
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    3/4 then okey

  7. anonymous
    • one year ago
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    I chose 3 as x2 and 0 and x1 same thing for y values. This should give me the new equation to y=3/4*x+ 25, Yes?

  8. MrNood
    • one year ago
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    no first - you have done x/y but the gradient is y/x so the gradient of the radius is 4/3 Now you need toknow that a tangent is always perpendicular to a radius at the point - so what do you know about gradient of perpendicular lines?

  9. anonymous
    • one year ago
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    sorry i didnt quite understand the last part, the book is in english but we are examined in swedish so there is a lot of translations I need to do. So u are asking me where the gradient of the lines meet?

  10. MrNood
    • one year ago
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    The slope of the radius is 4/3 But it asks for the equation of the tangent at that point. I have told you that a tangent is perpendicular to the radius (this is probably the key point for solving this question) So if the radius has slope 4/3 do you know haw to work out the slope of a line that is perpendicular to it?

  11. MrNood
    • one year ago
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    |dw:1434551549134:dw|

  12. anonymous
    • one year ago
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    No I do not know how to calculate the slope of that line...hmmm

  13. MrNood
    • one year ago
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    it is a basic property of straight lines. Can I ask you to research it - google it or refer to your book? There is a simple relationship between lines which are perpendicular. You will need to know it for other work on straight lines. Please post it when you have found it out :-)

  14. anonymous
    • one year ago
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    The thing is Im studying for calc 2, Ive dont everything properly, chapter discusses implicit differenttiation, suddenly an exercise like this shows up, I dont have any examples on this current chapter to back up on how to calculate such a thing, hence I dont know anything bout it, now atleast.

  15. MrNood
    • one year ago
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    If a line has slope m then a line which is perpendicular has slope -1/m so the radius has slope 4/3 what is the slope of the perpendicular tangent?

  16. anonymous
    • one year ago
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    -3/4

  17. MrNood
    • one year ago
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    OK good so now this is your question: |dw:1434552220146:dw| What is the equation of that line in y =mx+b form...?

  18. MrNood
    • one year ago
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    BTW - maths in th erela world does not come in 'subjects' this is not calc, but you should still be able to do straight lines

  19. anonymous
    • one year ago
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    y=-3/4x+25, got it, But Im really confused why they didnt bring up any examples or anything related to slopes and graphs, it was kind of a off chance exercise since its end of block part on the chapter.

  20. anonymous
    • one year ago
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    I will think about that, thanks for the help and ur patience :)

  21. MrNood
    • one year ago
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    y=-3/4x+25 I don't think tat is correct (the 25 is wrong)

  22. MrNood
    • one year ago
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    4 = -3 (3)/4 +b

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