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    the equation of the circle is  (x-3)^2 + (y-4)^2 = 25 , hence its centre is (3, 4) and radius 5cm.

    the second part of the question is :
    find the exact length of the tangents from the point (10, 0) to the circle.

    :confused: i don't even really understand the question.. doesn't the tangent always TOUCH the circle? and i don't know any points on the circumference of the circle to calculate the distance.

    the answer given is  2\sqrt10
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    (Original post by mandarinoriental)
    the equation of the circle is  (x-3)^2 + (y-4)^2 = 25 , hence its centre is (3, 4) and radius 5cm.

    the second part of the question is :
    find the exact length of the tangents from the point (10, 0) to the circle.

    :confused: i don't even really understand the question.. doesn't the tangent always TOUCH the circle? and i don't know any points on the circumference of the circle to calculate the distance.

    the answer given is  2\sqrt10
    Use Pythagorus' theorem. Have you drawn a diagram to show the circle, the point (10, 0) and the tangent. Now draw a line from the point (10, 0) to the centre of the circle and also draw a line from the centre of the circle to the point where the tangent meets the circle (i.e. a radius of the circle). Note that the radius to the circle that passes through the same point that the tangent does is perpendicular to the tangent i.e. the right angle is between the radius and the tangent. You know the length of the radius (from part (i)) and you can calculate the distance from (10, 0) to the centre of the circle.
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    Circles are round
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    (Original post by Farhan.Hanif93)
    Use Pythagorus' theorem. Have you drawn a diagram to show the circle, the point (10, 0) and the tangent. Now draw a line from the point (10, 0) to the centre of the circle and also draw a line from the centre of the circle to the point where the tangent meets the circle (i.e. a radius of the circle). Note that the radius to the circle that passes through the same point that the tangent does is perpendicular to the tangent i.e. the right angle is between the radius and the tangent. You know the length of the radius (from part (i)) and you can calculate the distance from (10, 0) to the centre of the circle.
    sorry but i dont understand. then can't i just use pythagorus' theorem straightaway, for example calculate the distance between (10, 0) and the centre straight away. why do i have to draw the radius? what happens after i draw the radius to the tangent?
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    (Original post by mandarinoriental)
    sorry but i dont understand. then can't i just use pythagorus' theorem straightaway, for example calculate the distance between (10, 0) and the centre straight away. why do i have to draw the radius? what happens after i draw the radius to the tangent?
    Yes, you can but I guessed that you were asking this question because you couldn't see how to solve the problem. That's why I suggested drawing it to make things clear. Things are a lot easier when you can see the triangle you're considering. Otherwise you'll be mindlessly plugging numbers into a formula, where you have no idea what the formula is being applied to. If you could visualise the problem then it's simply GCSE use of Pythagorus, in which case, it shouldn't be a problem at all.
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    Yes, you can but I guessed that you were asking this question because you couldn't see how to solve the problem. That's why I suggested drawing it to make things clear. Things are a lot easier when you can see the triangle you're considering. Otherwise you'll be mindlessly plugging numbers into a formula, where you have no idea what the formula is being applied to. If you could visualise the problem then it's simply GCSE use of Pythagorus, in which case, it shouldn't be a problem at all.
    I actually solved it calculating the distance straightaway but couldn't come to the correct answer..

    here's what i did (i calculated the distance between (10,0) and the centre (3,4)):

    \sqrt{(10-3)^2 + (0-4)^2}
    = \sqrt{7^2 + (-4)^2}
    = \sqrt65

    which cannot become and is not equivalent to the answer given, that is 2\sqrt 10
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    (Original post by mandarinoriental)
    I actually solved it calculating the distance straightaway but couldn't come to the correct answer..

    here's what i did (i calculated the distance between (10,0) and the centre (3,4)):

    \sqrt{(10-3)^2 + (0-4)^2}
    = \sqrt{7^2 + (-4)^2}
    = \sqrt65

    which cannot become and is not equivalent to the answer given, that is 2\sqrt 10
    I got the same answer, root 65.
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    (Original post by Farhan.Hanif93)
    Use Pythagorus' theorem. Have you drawn a diagram to show the circle, the point (10, 0) and the tangent. Now draw a line from the point (10, 0) to the centre of the circle and also draw a line from the centre of the circle to the point where the tangent meets the circle (i.e. a radius of the circle). Note that the radius to the circle that passes through the same point that the tangent does is perpendicular to the tangent i.e. the right angle is between the radius and the tangent. You know the length of the radius (from part (i)) and you can calculate the distance from (10, 0) to the centre of the circle.
    Good luck. I'll leave this one to you. If you can't persuade the OP to draw a sketch, you haven't got a chance!
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    (Original post by mandarinoriental)
    I actually solved it calculating the distance straightaway but couldn't come to the correct answer..

    here's what i did (i calculated the distance between (10,0) and the centre (3,4)):

    \sqrt{(10-3)^2 + (0-4)^2}
    = \sqrt{7^2 + (-4)^2}
    = \sqrt65

    which cannot become and is not equivalent to the answer given, that is 2\sqrt 10
    You haven't solved it because that is not the length of the tangent through (10, 0) to the circle, that's the distance between (10, 0) and the centre. This is why I suggested drawing that diagram, so you don't end up calculating the wrong distance like you did. You do need that distance to find the length of the tangent though. You now have to use pythagorus' theorem on the triangle I suggested in my first post. Don't forget that \sqrt{65} is the length of the hypotenuse of that triangle so you need to note that c^2-b^2=a^2 for a right angled triangle with sides of length a, b and c. Draw the diagram! :p:
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    (Original post by tiny hobbit)
    Good luck. I'll leave this one to you. If you can't persuade the OP to draw a sketch, you haven't got a chance!
    It's a little frustrating... :p:
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    thats because you calculated the distance to the centre which is not what the question asked for.
    I did this question a few weeks ago, listen to Farhan, he's right.
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    (Original post by Farhan.Hanif93)
    You haven't solved it because that is not the length of the tangent through (10, 0) to the circle, that's the distance between (10, 0) and the centre. This is why I suggested drawing that diagram, so you don't end up calculating the wrong distance like you did. You do need that distance to find the length of the tangent though. You now have to use pythagorus' theorem on the triangle I suggested in my first post. Don't forget that \sqrt{65} is the length of the hypotenuse of that triangle so you need to note that c^2-b^2=a^2 for a right angled triangle with sides of length a, b and c. Draw the diagram! :p:
    I GOT ITTTTT. thanks so much for your help and patience. i understand it now.

    (Original post by Farhan.Hanif93)
    It's a little frustrating... :p:
    yea sorryy had to make you explain so much. i'm not exactly very bright. especially when it comes to math. hahah

    thanks again!
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    yep, draw diagram, thanks farhan
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    (Original post by mandarinoriental)
    I GOT ITTTTT. thanks so much for your help and patience. i understand it now.

    yea sorryy had to make you explain so much. i'm not exactly very bright. especially when it comes to math. hahah

    thanks again!
    No worries.

    What I'll recommend is that whenever you come across a problem that you cannot just solve immediately, it's worthwhile drawing the situation out. It will get you out of trouble a lot of the time.
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    (Original post by Farhan.Hanif93)
    No worries.

    What I'll recommend is that whenever you come across a problem that you cannot just solve immediately, it's worthwhile drawing the situation out. It will get you out of trouble a lot of the time.
    I appreciate this tip.. Was stuck on this question as well. That sigh of relief when you finally get it. 😆
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    I got the same answer
 
 
 
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