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  • BITSAT Two circles x2 +y2-6x+8=0 and x2+y2-6 =0 are given. The equation of the circle through their point of intersection and the point (1,1

  1. x2+y2-3x+1=0

  2. x2+y2-x+5=0

  3. x2+y2-8x+6y=8

  4. x2+y2-4x+8y=7

Answers (1)

General equation of circle , x^{2}+y^{2}+2gx+2fy+c = 0

Now find the points of intersection of the two circles.

At the intersection, x_{1}^{2} +y_{1}^{2} -6x_{1} +8 = x_{1}^{2} +y_{1}^{2} -6            \Rightarrow -6x_{1} +8 = -6\Rightarrow x_{1} =\frac{7}{3}

Putting the value of x1 in the equation of any of the two circle to find y1.

\left ( \frac{7}{3} \right )^{2} +y_{1}^{2} - 6 = 0 \Rightarrow y_{1} = \frac{\sqrt{5}}{3} , y_{2} = -\frac{\sqrt{5}}{3}

Now we have 3 points through which the circle is passing, (x_{1},y_{1}) = (\frac{7}{3} , \frac{\sqrt{5}}{3})    ,  (x_{2},y_{2}) = (\frac{7}{3} , -\frac{\sqrt{5}}{3})     ,    (x_{3},y_{3}) = (1,1)

Now putting these points in general equation we will have 3 equations.

x_{1}^{2}+y_{1}^{2}+2gx_{1}+2fy_{1}+c = 0 .............(1)

x_{2}^{2}+y_{2}^{2}+2gx_{2}+2fy_{2}+c = 0...............(2)

x_{3}^{2}+y_{3}^{2}+2gx_{3}+2fy_{3}+c = 0..............(3)

Putting values and after solving these 3 equations we will get, g = \frac{-3}{2} ,   f = 0    ,   c = 1

Putting these values in the general equation we will get the equation of circle, x^{2} +y^{2} -3x +1 = 0

Option (1) is correct

 

 

Posted by

Abhishek Sahu

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