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Math Help - Conics: Quick Circle Question

  1. #1
    Raj
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    Conics: Quick Circle Question

    Determine the equation of the circle by the following information/transformation without graphing.

    a.) A circle with centre (2,-3) and passing through (1,4) is reflecred in the x-axis.

    So finding the centre is easy enough (x-2)^2 + (y-3)^2 = r^2

    How do i find the radius though? If the point it passes through had the same x-value as the center then it would be a radius of 7. But it doesn't so how do i find this out?

    Apparently the radius is square root of 50.
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  2. #2
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    Quote Originally Posted by Raj View Post
    Determine the equation of the circle by the following information/transformation without graphing.

    a.) A circle with centre (2,-3) and passing through (1,4) is reflecred in the x-axis.
    The first circle is (x-2)^2 + (y+3)^2 = 50.
    Its reflection in the x-axis is (x-2)^2 + (y-3)^2 = 50.
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  3. #3
    Raj
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    Quote Originally Posted by Plato View Post
    The first circle is (x-2)^2 + (y+3)^2 = 50.
    Its reflection in the x-axis is (x-2)^2 + (y-3)^2 = 50.
    HOW DO I FIND THE RADIUS. I know this already, ugh.

    edit: nvm realized i just substitue the numbers of the point into the circle equation -.-
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  4. #4
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    Quote Originally Posted by Raj View Post
    HOW DO I FIND THE RADIUS.
    Oh my goodness. Get real.
    The radius of any circle is the distance from the center to any point on the circle.
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  5. #5
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    Quote Originally Posted by Plato View Post
    Oh my goodness. Get real.
    The radius of any circle is the distance from the center to any point on the circle.
    You initially got the equation of the circle wrong.

    The equation is (X-2)sqr + (y+3)sqr = Rsqr-------------(1

    But since the circle passes through (1,4) sub it into equation (1

    so you get

    (1-2)sqr + (4+3)sqr = Rsqr

    1 + 7sqr = Rsqr

    50 = Rsqr

    R = 5Root 2

    *** sqr = Squared

    Always remember that if a line or any shape passes through a point, the equation must be satisfied when you sub that point into that equation. like what i did above

    hope it helped

    ask me if u have any further questions
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  6. #6
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    Quote Originally Posted by strik3r View Post
    You initially got the equation of the circle wrong.

    The equation is (X-2)sqr + (y+3)sqr = Rsqr-------------(1

    But since the circle passes through (1,4) sub it into equation (1

    so you get

    (1-2)sqr + (4+3)sqr = Rsqr

    1 + 7sqr = Rsqr

    50 = Rsqr

    R = 5Root 2

    *** sqr = Squared

    Always remember that if a line or any shape passes through a point, the equation must be satisfied when you sub that point into that equation. like what i did above

    hope it helped

    ask me if u have any further questions
    I don't see why you've quoted Plato and then posted what you have posted. Plato has not made any mistake.
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