Conic Sections PreCalculus

  • Conic Sections PreCalculus

  • lLammar Academy

    lLammar Academy

    Organizer
    April 14, 2021 at 12:20 pm

    What is the standard equation of a circle?

    (x-\blueD h)^2+(y-\maroonD k)^2=\goldD r^2(x−h)2+(y−k)2=r2left parenthesis, x, minus, start color #11accd, h, end color #11accd, right parenthesis, squared, plus, left parenthesis, y, minus, start color #ca337c, k, end color #ca337c, right parenthesis, squared, equals, start color #e07d10, r, end color #e07d10, squared

    This is the general standard equation for the circle centered at (\blueD h, \maroonD k)(h,k)left parenthesis, start color #11accd, h, end color #11accd, comma, start color #ca337c, k, end color #ca337c, right parenthesis with radius \goldD rrstart color #e07d10, r, end color #e07d10.

    Circles can also be given in expanded form, which is simply the result of expanding the binomial squares in the standard form and combining like terms.

    For example, the equation of the circle centered at (\blueD 1,\maroonD 2)(1,2)left parenthesis, start color #11accd, 1, end color #11accd, comma, start color #ca337c, 2, end color #ca337c, right parenthesis with radius \goldD 33start color #e07d10, 3, end color #e07d10 is (x-\blueD 1)^2+(y-\maroonD 2)^2=\goldD 3^2(x−1)2+(y−2)2=32left parenthesis, x, minus, start color #11accd, 1, end color #11accd, right parenthesis, squared, plus, left parenthesis, y, minus, start color #ca337c, 2, end color #ca337c, right parenthesis, squared, equals, start color #e07d10, 3, end color #e07d10, squared. This is its expanded equation:

    \begin{aligned} (x-\blueD 1)^2+(y-\maroonD 2)^2&=\goldD 3^2 \\\\ (x^2-2x+1)+(y^2-4y+4)&=9 \\\\ x^2+y^2-2x-4y-4&=0 \end{aligned}(x−1)2+(y−2)2(x2−2x+1)+(y2−4y+4)x2+y2−2x−4y−4​=32=9=0​

    Want to learn more about circle equations? Check out this video.

    Practice set 1: Using the standard equation of circles

    PROBLEM 1.1

    (x+4)^{2}+(y-6)^{2} = 48(x+4)2+(y−6)2=48left parenthesis, x, plus, 4, right parenthesis, squared, plus, left parenthesis, y, minus, 6, right parenthesis, squared, equals, 48

    What is the center of the circle?

    ((left parenthesis

    ,,comma ))right parenthesis

    What is its radius?
    If necessary, round your answer to two decimal places.

    units

    Want to try more problems like this? Check out this exercise and this exercise.

    Practice set 2: Writing circle equations

    PROBLEM 2.1

    A circle has a radius of \sqrt{13}13​square root of, 13, end square root units and is centered at (-9.3,4.1)(−9.3,4.1)left parenthesis, minus, 9, point, 3, comma, 4, point, 1, right parenthesis.

    Write the equation of this circle.

    Want to try more problems like this? Check out this exercise.

    Practice set 3: Using the expanded equation of circles

    To interpret the expanded equation of a circle, we should rewrite it in standard form using the method of “completing the square.”

    Consider, for example, the process of rewriting the expanded equation x^2+y^2+18x+14y+105=0x2+y2+18x+14y+105=0x, squared, plus, y, squared, plus, 18, x, plus, 14, y, plus, 105, equals, 0 in standard form:

    x^2+y^2+18x+14y+105&=0

    x^2+y^2+18x+14y&=-105

    (x^2+18x)+(y^2+14y)&=-105

    (x^2+18x\redD{+81})+(y^2+14y{+49})&=-105\redD{+81}\blueD{+49}

    (x+\redD9)^2+(y+\blueD7)^2&=25 \\\\ (x-(-9))^2+(y-(-7))^2&=5^2 \end{aligned}x2+y2+18x+14y+105×2+y2+18x+14y(x2+18x)+(y2+14y)(x2+18x+81)+(y2+14y+49)(x+9)2+(y+7)2(x−(−9))2+(y−(−7))2​=0=−105=−105=−105+81+49=25=52​

    Now we can tell that the center of the circle is (-9,-7)(−9,−7)left parenthesis, minus, 9, comma, minus, 7, right parenthesis and the radius is 555.

    PROBLEM 3.1

    x^2+y^2-10x-16y+53=0x2+y2−10x−16y+53=0x, squared, plus, y, squared, minus, 10, x, minus, 16, y, plus, 53, equals, 0

    What is the centre of this circle?

    What is the radius of this circle/\?

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