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Exercise 12.2 · Q7

Q.Find the value of kk for which the circles x2+y2−3x+ky−5=0x^2 + y^2 - 3x + ky - 5 = 0 and 4x2+4y2−12x−y−9=04x^2 + 4y^2 - 12x - y - 9 = 0 are concentric.

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Concentric circles share the same centre; equate the centres of the two given circles (after normalising each to the form with unit x2,y2x^2,y^2 coefficients).

Centre of x2+y2+2gx+2fy+c=0x^2+y^2+2gx+2fy+c=0 is (−g,−f)(-g,-f).

  1. First circle: x2+y2−3x+ky−5=0x^2+y^2-3x+ky-5=0. Here 2g1=−3⇒g1=−322g_1=-3\Rightarrow g_1=-\dfrac32; 2f1=k⇒f1=k22f_1=k\Rightarrow f_1=\dfrac k2. Centre1=(32,−k2)_1=\left(\dfrac32,-\dfrac k2\right).
  2. Second circle: 4x2+4y2−12x−y−9=04x^2+4y^2-12x-y-9=0. Divide by 4 to normalise: x2+y2−3x−14y−94=0x^2+y^2-3x-\dfrac14y-\dfrac94=0. Here 2g2=−3⇒g2=−322g_2=-3\Rightarrow g_2=-\dfrac32; 2f2=−14⇒f2=−182f_2=-\dfrac14\Rightarrow f_2=-\dfrac18. Centre2=(32,18)_2=\left(\dfrac32,\dfrac18\right). …

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