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Math Help - Prove algebraically- the values of k that do no intersect the circle

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    Prove algebraically- the values of k that do no intersect the circle

    Hope im in the right forum post
    Basically i have to prove algebraically- the values of k in y=kx+7 that do no intersect the circle x2 +y2=25

    What i already know -In my text book i learnt that the radius of the circle is sqrt 25=5
    so radius all the way around the circle is 5.

    The line y=kx+7 has a y intercept of 7, where it cuts the y axis.

    Im not sure about how to go on proving this algebraically, i read an article that you have to sustitute the line equation into the circle one on this site :BBC - Higher Bitesize Maths - The circle : Page 5
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    Re: Prove algebraically- the values of k that do no intersect the circle

    oh yeah and k would be the gradient
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    Re: Prove algebraically- the values of k that do no intersect the circle

    Quote Originally Posted by studenthelp10 View Post
    Hope im in the right forum post
    Basically i have to prove algebraically- the values of k in y=kx+7 that do no intersect the circle x2 +y2=25
    Find the values of k for which x^2+(kx+7)^2=25 has no real roots.
    Thanks from studenthelp10
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    Re: Prove algebraically- the values of k that do no intersect the circle

    Yes, to investigate intersections sub y=kx+7 into x^2+y^2=25

    So x^2+(kx+7)^2=25 x^2+k^2x^2+14kx+49-25=0

    Arrange as a quadratic in x Get (1+k^2)x^2+14kx+24=0

    We want this to have no real roots. (The condition for ax^2+bx+c=0 to have no real roots is b^2-4ac<0). Try and apply this here.
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    Re: Prove algebraically- the values of k that do no intersect the circle

    thanks guys ill try that out now!!! your the best!
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    Re: Prove algebraically- the values of k that do no intersect the circle

    is this applying it ?
    a b c
    b^2-4ac<0 then sub into it (1+k^2)x^2+14kx+24=0

    making it :
    14^2-4(1+k^2)(24)<0
    196- 4+4k^2(24)<0
    196- 96 +96k^2<0
    100+96k^2<0
    96k^2<-100
    k^2<-100/96
    k<sqrt (-100/96)

    Am i making progress? or have i done it completely wrong?
    also why do we want it to have no real roots ? what do real roots do?
    Last edited by studenthelp10; April 15th 2012 at 04:49 AM.
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    Re: Prove algebraically- the values of k that do no intersect the circle

    but then what are all the values that miss the circle - from the closest possible to the circle , then outward for 'k' going back to my main question. K is the gradient that controls the slope of the line - so how do i find all the values of k that do not intersect the circle given the equation of the line is y=kx+7 and the circle is x^2 + y^2=25.
    Last edited by studenthelp10; April 15th 2012 at 04:52 AM.
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    Re: Prove algebraically- the values of k that do no intersect the circle

    You are using the correct method but your 3rd line should have been (14k)^2-4(1+k)24<0 Yo left out the 1st k.

    If there were some roots these would be the x coordinates of the points of intersection. So no roots means no intersections.
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    Re: Prove algebraically- the values of k that do no intersect the circle

    Thank biffboy for the reply - much appreciated
    i probably worded a bit wrong at the start but i wanted to find out all the values of k that do not intersect the circle, given that k is the gradient- changing the slope of the line . Im sure its gonna be an answer like all values of must k < ? to not intersect the circle. But i need to prove it algebraically. would the above answer be on the right track to solving the problem? If not could you show me how or hint me towards what i need to do to get the answer thanks so much

    made the change
    (14k)^2-4(1+k^2)(24)<0
    196k- 4+4k^2(24)<0
    196k- 96 +96k^2<0
    96k^2+196k+96<0 (rearranging)
    (k )(k )
    or if cant find 2 factors use quadratic formula




    ill post to you tommorrow coz i need to get some rest- thanks for the help so far
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    Re: Prove algebraically- the values of k that do no intersect the circle

    There's still an error. Note that -4(1+k^2)(24)=-96(1+k^2)=-96-96k^2
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    Re: Prove algebraically- the values of k that do no intersect the circle

    i think because "k" is any value that does not make the line intersect/ cross with the circle- thats probably why theres and error - still need to find the range of values that for k that does not cross the circle

    (14k)^2-4(1+k^2)(24)<0

    196k-96(1+k^2)<0
    196k-96-96k^2 <0
    -96k^2+196k-96 <0(rearranging)
    better?
    then factorise or use quadratic formula?
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    Re: Prove algebraically- the values of k that do no intersect the circle

    In your 2nd line that 196k should be 196k^2

    So 196k^2-96-96k^2<0 So 100k^2<96 So k^2<96/100 So K^2<0.96 So -root 0.96<x<+root0.96

    Note for root(96/100) the book has probably done these steps. root(96/100)=root96/root100=root(16x6)/10=(4root6)/10 =(2root6)/5
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    Re: Prove algebraically- the values of k that do no intersect the circle

    Thanks again biffboy! im grateful for youre help

    im making simple mistakes - but thanks for correcting me.
    so once i get the answer would all the range of values for k be -root 0.96<x<+root0.96 meaning between -root0.96 and +root0.96 would be all the values for k- the gradient?

    Remembering the question was "Find the range of values for "k" , for which the circle x^2 + y^2=25 and the line y=kx+7 do not intersect"
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    Re: Prove algebraically- the values of k that do no intersect the circle

    That's correct.
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    Re: Prove algebraically- the values of k that do no intersect the circle

    oh thanks for the help ^_^' that problem wasnt to bad afterall after the explanation " thank you !
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