The equation of the circle which passes through the points of intersection of the circles $x^{2} + y^{2} - 6 x = 0$ and $x^{2} + y^{2} - 6 y = 0$ and has its centre at $(\frac{3}{2}, \frac{3}{2})$ , is

x^{2} + y^{2} + 3 x + 3 y + 9 = 0

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x^{2} + y^{2} + 3 x + 3 y = 0

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x^{2} + y^{2} - 3 x - 3 y = 0

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x^{2} + y^{2} - 3 x - 3 y + 9 = 0

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Solution

The correct option is C $x^{2} + y^{2} - 3 x - 3 y = 0$
The equation of given circles are
$x^{2} + y^{2} - 6 x = 0$ .....(i)
and $x^{2} + y^{2} - 6 y = 0$ .....(ii)
On solving Eqs. (i) and (ii), we get
$x = 0, y = 0$ or $x = 3, y = 3$
$∴$ Points of intersection are (0, 0) and (3, 3).
Also, centre of required circle is $(\frac{3}{2}, \frac{3}{2})$ .
$∴ g = - \frac{3}{2}$ and $f = - \frac{3}{2}$
Hence, equation of circle is
$x^{2} + y^{2} - 3 x - 3 y + c = 0$
Since, this circle passes through (0, 0), thus equation of circle becomes
$x^{2} + y^{2} - 3 x - 3 y = 0$

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