Q

# 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

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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

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