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    Find the equation of the tangent to the circle (x-3)^2+(y+2)^=125 at thepoint (-2, 8).

    Find the coordinates of the points where the circle (x-3)^2+(y-2)^2=40 cutsthe x-axis.

    Find the coordinates of the points where the line y = x + 4 meets the circle
    Equation of circle: (x-3)^2+(y-5)^2 = 34

    How would you do this? Help pls
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    have you made any start?
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    (Original post by CookieHero)
    Find the equation of the tangent to the circle (x-3)^2+(y+2)^=125 at thepoint (-2, 8).

    Find the coordinates of the points where the circle (x-3)^2+(y-2)^2=40 cutsthe x-axis.

    Find the coordinates of the points where the line y = x + 4 meets the circle
    Equation of circle: (x-3)^2+(y-5)^2 = 34

    How would you do this? Help pls
    1. The centre of the circle has coordinates (3, -2)
    You now have two sets of coordinates. Using these; (3, -2) & (-2, 8); find the gradient of the line connecting these two points. Then, you can work out the gradient of the tangent (negative reciprocal), which is perpendicular to that line.
    Using y=mx+c, substitute your coordinates in and the gradient to find the value of C. And you have your equation of the tangent!

    2. Remember... When a line, curve or circle crosses the x axis, y is equal to 0. So substitute y=0 and work out the values of x from there.

    3. It's simultaneous equations. You know at the points where the lines meet they both equal one another.
    The best way to go about this is by substitution.
    Replace all the y values in your circle equation with x+4; because y=x+4.

    The details should be trivial. If they aren't, make sure you check with your teacher ASAP for help!
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    (Original post by Jamdroid)
    1. The centre of the circle has coordinates (3, -2)
    You now have two sets of coordinates. Using these; (3, -2) & (-2, 8); find the gradient of the line connecting these two points. Then, you can work out the gradient of the tangent (negative reciprocal), which is perpendicular to that line.
    Using y=mx+c, substitute your coordinates in and the gradient to find the value of C. And you have your equation of the tangent!

    2. Remember... When a line, curve or circle crosses the x axis, y is equal to 0. So substitute y=0 and work out the values of x from there.

    3. It's simultaneous equations. You know at the points where the lines meet they both equal one another.
    The best way to go about this is by substitution.
    Replace all the y values in your circle equation with x+4; because y=x+4.

    The details should be trivial. If they aren't, make sure you check with your teacher ASAP for help!
    Thanks for the advice but i still have some issues.

    For the first question I keep getting y = 1/2x+9 when in the solutions it says y = -0.5x+7. Working out-2-8)/(3+2) = -2m1 = -2 therefore m2 = 1/2y-8 = 1/2(x+2)y=1/2x+1+8y=1/2x + 9how is this wrong???

    For the second question I got (9,0) which was right but I cant seem to figure out how to to get (-3,0) which says in the answers.

    For the third question, im still doing it
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    (Original post by CookieHero)
    Thanks for the advice but i still have some issues.

    For the first question I keep getting y = 1/2x+9 when in the solutions it says y = -0.5x+7. Working out-2-8)/(3+2) = -2m1 = -2 therefore m2 = 1/2y-8 = 1/2(x+2)y=1/2x+1+8y=1/2x + 9how is this wrong???

    For the second question I got (9,0) which was right but I cant seem to figure out how to to get (-3,0) which says in the answers.

    For the third question, im still doing it
    Nvm, I solved the 2nd question. Its just the 1st question thats bugging me now.
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    (Original post by CookieHero)
    Thanks for the advice but i still have some issues.

    For the first question I keep getting y = 1/2x+9 when in the solutions it says y = -0.5x+7. Working out-2-8)/(3+2) = -2m1 = -2 therefore m2 = 1/2y-8 = 1/2(x+2)y=1/2x+1+8y=1/2x + 9how is this wrong???

    For the second question I got (9,0) which was right but I cant seem to figure out how to to get (-3,0) which says in the answers.

    For the third question, im still doing it
    For the first question, I can't see a flaw in your logic. Are you sure you wrote the question & coordinates down correctly?

    In the second question, you should end up with a quadratic equation. When you factorise you will (usually) get two solutions. If this is not the case, again you should just check you wrote the question correctly, and if you're checking the right answers!
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    It seems like the mark scheme is wrong and thanks da help
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    (Original post by CookieHero)
    It seems like the mark scheme is wrong and thanks da help
    You're most welcome!
 
 
 
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