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The equation of a circle x2 + y2 − 2ax + a2 + 4y = 0. a) centre b) area of triangle in terms of centre, r2. x2 + y2 − 2bx + 4y + b2 = 1

Mathematics

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2 parts
a

(32,−2)\left(\frac{3}{2}, -2\right)

(1.5, -2)
b

254π\frac{25}{4}\pi

Step-by-step solution

5 steps

Part a· 2 steps

Kituo cha duara.

Equation ya duara iliyotolewa ni x2+y2−3x+4y=0x^2 + y^2 - 3x + 4y = 0.
Equation ya jumla ya duara ni x2+y2+2gx+2fy+c=0x^2 + y^2 + 2gx + 2fy + c = 0, ambapo kituo cha duara ni (−g,−f)(-g, -f).

  1. Linganisha coefficients za xx na yy kutoka equation uliyopewa na equation ya jumla.

    2gx=−3x  ⟹  2g=−3  ⟹  g=−322gx = -3x \implies 2g = -3 \implies g = -\frac{3}{2} 2fy=4y  ⟹  2f=4  ⟹  f=22fy = 4y \implies 2f = 4 \implies f = 2
  2. Tumia maadili ya gg na ff kupata kituo cha duara.

    Kituo ni (−g,−f)(-g, -f).

    (−g,−f)=(−(−32),−2)=(32,−2)(-g, -f) = \left(-\left(-\frac{3}{2}\right), -2\right) = \left(\frac{3}{2}, -2\right)

    Au kwa desimali: (1.5,−2)(1.5, -2).

    Kituo cha duara ni (32,−2)\boxed{\left(\frac{3}{2}, -2\right)} au (1.5,−2)\boxed{(1.5, -2)}.

Part b

254π\frac{25}{4}\pi

Eneo la duara.

Kuna uwezekano kwamba "area of triangle" ni makosa ya kuandika na inapaswa kuwa "area of circle", hasa kwa sababu swali linaomba jibu "in terms of π\pi". Tutachukulia kuwa swali linauliza eneo la duara. Equation iliyotolewa katika sehemu b) ni x2−2ax+a2+y2−2by+b2=r2x^2 - 2ax + a^2 + y^2 - 2by + b^2 = r^2, ambayo inaweza kuandikwa kama (x−a)2+(y−b)2=r2(x-a)^2 + (y-b)^2 = r^2. Hii ni equation ya duara yenye kituo (a,b)(a,b) na radius rr. Tutatumia duara ileile kutoka sehemu a).

  1. Andika upya equation ya duara kutoka sehemu a) katika fomu ya kawaida (x−a)2+(y−b)2=r2(x-a)^2 + (y-b)^2 = r^2 kwa kukamilisha mraba.

    Equation: x2+y2−3x+4y=0x^2 + y^2 - 3x + 4y = 0
    Panga upya maneno ya xx na yy:

    (x2−3x)+(y2+4y)=0(x^2 - 3x) + (y^2 + 4y) = 0

    Kamilisha mraba kwa maneno ya xx na yy:

    (x2−3x+(−32)2)+(y2+4y+(42)2)=(−32)2+(42)2\left(x^2 - 3x + \left(-\frac{3}{2}\right)^2\right) + \left(y^2 + 4y + \left(\frac{4}{2}\right)^2\right) = \left(-\frac{3}{2}\right)^2 + \left(\frac{4}{2}\right)^2 (x−32)2+(y+2)2=94+4\left(x - \frac{3}{2}\right)^2 + (y + 2)^2 = \frac{9}{4} + 4 (x−32)2+(y+2)2=94+164\left(x - \frac{3}{2}\right)^2 + (y + 2)^2 = \frac{9}{4} + \frac{16}{4} (x−32)2+(y+2)2=254\left(x - \frac{3}{2}\right)^2 + (y + 2)^2 = \frac{25}{4}
  2. Tambua r2r^2 kutoka equation iliyokamilishwa.

    Kutoka equation (x−1.5)2+(y+2)2=254(x - 1.5)^2 + (y + 2)^2 = \frac{25}{4}, tunaona kwamba r2=254r^2 = \frac{25}{4}.

  3. Kokotoa eneo la duara.

    Eneo la duara ni A=πr2A = \pi r^2.

    A=π(254)A = \pi \left(\frac{25}{4}\right) A=254πA = \frac{25}{4}\pi

    Eneo la duara ni 254π\boxed{\frac{25}{4}\pi}.

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