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    The centre of a circle is (4,0). The line 2x+y=3 is a tangent to the circle. Find the equation of the circle.

    So I've racked my brains trying to handle this question, the first thing I did was
    put the tangent into slope intercept form : y=-2x+3
    and I know that the tangent is perpendicular to the radius so i figured out the equation of the radius where it meets the tangent which is: y=1/2 x-2.
    Then, I tried to make the circle equation:
    (x-4)^2+ y^2= r^2
    however I don't know how to figure out what the radius is, help would be really appreciated
    Here is the question along with the diagram Attachment 618776
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    (Original post by TeeEm)
    ...call the radius squared as c for simplicity
    Solve the equation simultaneously
    Get a quadratic in x or y where c is contained in the coefficients
    Look for repeated roots ...
    Oh! Thank you, I'll try it out right now
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    (Original post by TeeEm)
    ...call the radius squared as c for simplicity
    Solve the equation simultaneously
    Get a quadratic in x or y where c is contained in the coefficients
    Look for repeated roots ...
    Wait sorry, which equation would i substitute in for y?
    y=-2x+3 or y=1/2x-2
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    (Original post by TeeEm)
    y=-2x+3
    Sorry to keep bothering you but I did this:
    (x-4)^2 + (-2x+3)^2=c
    and all expanded and simplified gives me:
    x^2-4x+5 which factorises to (x+1)(x-5)
    I have two x values now but I'm not sure what to do next..
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    (Original post by Mermaidqueen)
    Sorry to keep bothering you but I did this:
    (x-4)^2 + (-2x+3)^2=c
    and all expanded and simplified gives me:
    x^2-4x+5 which factorises to (x+1)(x-5)
    I have two x values now but I'm not sure what to do next..
    That's not quite right. You need to get the form Ax^2+Bx+C=0 then consider the discriminant and what the condition must be upon it for the line to be tangent (if the line is tangent, how many roots should this equation have? - ie points of intersection), where C is in terms of c, and then use the condition on the discriminant to solve for c
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    (Original post by RDKGames)
    That's not quite right. You need to get the form Ax^2+Bx+C=0 then consider the discriminant and what the condition must be upon it for the line to be tangent (if the line is tangent, how many roots should this equation have? - ie points of intersection), where C is in terms of c, and then use the condition on the discriminant to solve for c
    In this case c represents r squared, I'm a bit confused.. I understand that if the line is a tangent there can only be one root, however I'm quite stuck
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    (Original post by Mermaidqueen)
    In this case c represents r squared, I'm a bit confused.. I understand that if the line is a tangent there can only be one root, however I'm quite stuck
    Alright, so you have (x-4)^2+(-2x+3)^2=r^2

    Fully expand the LHS, collect all the common terms and get it in the form Ax^2+Bx+C=0 then apply the discriminant.
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    (Original post by RDKGames)
    Alright, so you have (x-4)^2+(-2x+3)^2=r^2

    Fully expand the LHS, collect all the common terms and get it in the form Ax^2+Bx+C=0 then apply the discriminant.
    By fully expanding the LHS and collecting the like terms and simplifying it
    I got x^2-4x+5= r^2

    I searched to find out what a discriminant is and I still don't know, I'm really sorry
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    (Original post by Mermaidqueen)
    By fully expanding the LHS and collecting the like terms and simplifying it
    I got x^2-4x+5= r^2

    I searched to find out what a discriminant is and I still don't know, I'm really sorry
    (x^2-8x+16)+(4x^2-12x+9)=r^2 \Rightarrow 5x^2-20x+(25-r^2)=0

    The discriminant is b^2-4ac for a quadratic in the form ax^2+bx+c=0
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    (Original post by Mermaidqueen)
    The centre of a circle is (4,0). The line 2x+y=3 is a tangent to the circle. Find the equation of the circle.

    So I've racked my brains trying to handle this question, the first thing I did was
    put the tangent into slope intercept form : y=-2x+3
    and I know that the tangent is perpendicular to the radius so i figured out the equation of the radius where it meets the tangent which is: y=1/2 x-2.
    Then, I tried to make the circle equation:
    (x-4)^2+ y^2= r^2
    however I don't know how to figure out what the radius is, help would be really appreciated
    Here is the question along with the diagram Attachment 618776
    The included diagram makes me think you're supposed to take a more 'geometrical' approach, rather than substitutuing into the circle equation and looking for repeated roots.

    The easiest way to solve this (assuming you know the formula for the distance between a point and a line) is to observe that the radius is simply the distance between the line y=-2x+3 and the point (4, 0).

    Alternatively (and again, following what the diagram hints at): Find the equation of the other line in the diagram. That is, the line perpendicular to y=-2x+3 that passes through (4, 0). Find where this line intersects y=-2x+3. This gives you a point on the circumference and you can find the radius.
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    (Original post by DFranklin)
    The included diagram makes me think you're supposed to take a more 'geometrical' approach, rather than substitutuing into the circle equation and looking for repeated roots.

    The easiest way to solve this (assuming you know the formula for the distance between a point and a line) is to observe that the radius is simply the distance between the line y=-2x+3 and the point (4, 0).

    Alternatively (and again, following what the diagram hints at): Find the equation of the other line in the diagram. That is, the line perpendicular to y=-2x+3 that passes through (4, 0). Find where this line intersects y=-2x+3. This gives you a point on the circumference and you can find the radius.
    Thank you very much!!! I got the answer and I completely understood it.
    (my final answer was (x-4)^2+y^2=5 in case you were wondering
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    (Original post by RDKGames)
    (x^2-8x+16)+(4x^2-12x+9)=r^2 \Rightarrow 5x^2-20x+(25-r^2)=0

    The discriminant is b^2-4ac for a quadratic in the form ax^2+bx+c=0
    I got to the answer in the end, thank you very much for helping!
 
 
 
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