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    Hi all!

    I'm a bit stuck on the following two questions and was wondering whether anybody could give me a hand?

    5. Find the equation of the tangents to the circle x^2 + y^2+4x-2y-24=0 at the points where the circle cuts the line y=x.

    I substituted x in every time there was a y in the circle equation. Is this right to do?

    6. Show that the circles x^2 + y^2 -10x-8y+18 and x^2 + y^2 - 8x -4y+14=0 don't intersect.

    I put these two equal to each other as they both equal 0 and then simplified before finding the discriminant using D=b^2-4ac

    Thanks in advance for any help
    Blake
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    (Original post by Blake Jones)
    Hi all!

    I'm a bit stuck on the following two questions and was wondering whether anybody could give me a hand?

    5. Find the equation of the tangents to the circle x^2 + y^2+4x-2y-24=0 at the points where the circle cuts the line y=x.

    I substituted x in every time there was a y in the circle equation. Is this right to do?
    Yes? That'll give you the point of tangency, but you still need to find the equation of the tangent at that point.

    6. Show that the circles x^2 + y^2 -10x-8y+18 and x^2 + y^2 - 8x -4y+14=0 don't intersect.

    I put these two equal to each other as they both equal 0 and then simplified before finding the discriminant using D=b^2-4ac

    Thanks in advance for any help
    Blake
    I'm unsure as to how you used the discriminant since once you equate them and simplify, the x^2 and y^2 cancel out and it's no longer a quadratic equation, but is instead a linear equation.
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    (Original post by Zacken)
    Yes? That'll give you the point of tangency, but you still need to find the equation of the tangent at that point.

    I'm unsure as to how you used the discriminant since once you equate them and simplify, the x^2 and y^2 cancel out and it's no longer a quadratic equation, but is instead a linear equation.
    Pt1 When I did this I got x^2+x-14=0 meaning x=(-1+/-sqrt57)/2 Did you get this too or is this where my mistake has been made?

    Pt2 Ah, that's probably where I've gone wrong then... I'll try it again!
    x^2 + y^2 -10x-8y+18 = x^2 + y^2 - 8x -4y+14 I think it must have been my bad handwriting! Oops!
    -10x-8y+18=14-8x-4y
    -2x+4=4y
    y=1-1/2x I'm not sure how to show it from this point onwards though

    Thanks,
    Blake
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    (Original post by Blake Jones)
    Pt1 When I did this I got x^2+x-14=0 meaning x=(-1+/-sqrt57)/2 Did you get this too or is this where my mistake has been made?
    It is: You should get: x^2 + x^2 + 4x - 2x - 24 = 0 \iff 2x^2 + 2x - 24 = 0 \iff x^2 + x - 12 = 0.

    i.e: the constant term should be 12 not 14.

    Pt2 Ah, that's probably where I've gone wrong then... I'll try it again!
    x^2 + y^2 -10x-8y+18 = x^2 + y^2 - 8x -4y+14 I think it must have been my bad handwriting! Oops!
    -10x-8y+18=14-8x-4y
    -2x+4=4y
    y=1-1/2x I'm not sure how to show it from this point onwards though

    Thanks,
    Blake
    Okay, so what you have so far is "the circles intersect if and only if it is true that y = 1 - x/2".

    Now try plugging this y=1-x/2 into one of your circle equations to get an equation in either only x (or you can sub in x = 2-2y to get an equation in only y) and show that this produces an equation that is not solvable (you could consider the discriminant there).
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    (Original post by Zacken)
    It is: You should get: x^2 + x^2 + 4x - 2x - 24 = 0 \iff 2x^2 + 2x - 24 = 0 \iff x^2 + x - 12 = 0.

    i.e: the constant term should be 12 not 14.



    Okay, so what you have so far is "the circles intersect if and only if it is true that y = 1 - x/2".

    Now try plugging this y=1-x/2 into one of your circle equations to get an equation in either only x (or you can sub in x = 2-2y to get an equation in only y) and show that this produces an equation that is not solvable (you could consider the discriminant there).
    Thank you so much! I think working from 6am is starting to take its toll on me! I reckon it's time for a break!
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    (Original post by Zacken)
    It is: You should get: x^2 + x^2 + 4x - 2x - 24 = 0 \iff 2x^2 + 2x - 24 = 0 \iff x^2 + x - 12 = 0.

    i.e: the constant term should be 12 not 14.
    I've just been carrying on with this question and I have got the x and y values but bearing in mind I'm trying to find the equation of the tangent how do I go about finding the gradient? I have the centre of the circle but no other coordinates.

    Thanks again
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    (Original post by Blake Jones)
    I've just been carrying on with this question and I have got the x and y values but bearing in mind I'm trying to find the equation of the tangent how do I go about finding the gradient? I have the centre of the circle but no other coordinates.

    Thanks again
    That's not true, you have the centre and you know that the point on the circumference is (3,3) (from solving your quadratic and you know y=x) and also (-4, -4). (assuming I've solved the quadratic right).

    Then you can easily find the gradient of the radius and GCSE circle theorems tells you the gradient of the tangent is perpendicular to the gradient of the radius, so gradient of tangent = -1/(gradient of radius).
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    (Original post by Zacken)
    That's not true, you have the centre and you know that the point on the circumference is (3,3) (from solving your quadratic and you know y=x) and also (-4, -4). (assuming I've solved the quadratic right).

    Then you can easily find the gradient of the radius and GCSE circle theorems tells you the gradient of the tangent is perpendicular to the gradient of the radius, so gradient of tangent = -1/(gradient of radius).
    Ah yes I see! I forgot the one on the circumference!
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    (Original post by Blake Jones)
    Ah yes I see! I forgot the one on the circumference!
    Is it all sorted now? :-)
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    (Original post by Zacken)
    Is it all sorted now? :-)
    Yep, thank you for all your help! Much appreciated!
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    (Original post by Blake Jones)
    Yep, thank you for all your help! Much appreciated!
    No worries!
 
 
 
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