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Exercise 12.2 · Q4

Q.One end of a diameter of the circle x2+y2−6x+5y−7=0x^2 + y^2 - 6x + 5y - 7 = 0 is (−1,3)(-1, 3). Find the coordinates of the other end of the diameter.

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The centre of a circle is the midpoint of any diameter; find the centre from the equation, then use the midpoint formula.

Centre of x2+y2+2gx+2fy+c=0x^2+y^2+2gx+2fy+c=0 is (−g,−f)(-g,-f). Midpoint formula: centre =(x1+x22,y1+y22)=\left(\dfrac{x_1+x_2}{2},\dfrac{y_1+y_2}{2}\right).

  1. Given circle x2+y2−6x+5y−7=0x^2+y^2-6x+5y-7=0: match 2g=−6⇒g=−32g=-6\Rightarrow g=-3; 2f=5⇒f=2.52f=5\Rightarrow f=2.5.
  2. Centre =(−g,−f)=(3,−2.5)=(-g,-f)=(3,-2.5).
  3. One end of the diameter is (−1,3)(-1,3). Let the other end be (x2,y2)(x_2,y_2). Since the centre is the midpoint: (−1+x22,3+y22)=(3,−2.5)\left(\frac{-1+x_2}{2},\frac{3+y_2}{2}\right)=(3,-2.5) …

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