A circle passes through the origin anf has its centre on y = x. If it cuts x² + y² - 4x - 6y + 10 = 0 orthogonally, the equation of the circle is

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$$\begin{gathered} Let the equation of circle be x^2+y^2+2gx+2hy+f=0. \\ Therefore centre of circle is \left(-g,-h\right). \\ Since centre lies on line y=x, \therefore -h=-g \\ \Rightarrow h=g \\ Thus equation of circle becomes x^2+y^2+2gx+2gy+f=0. \\ Since the circle passes through origin, \\ \Rightarrow 0+0+2g\left(0\right)+2g\left(0\right)+f=0 \\ \Rightarrow f=0 \\ Hence, equation of circle becomes x^2+y^2+2gx+2gy=0. \\ Since it cuts x^2+y^2-4x-6y+10=0 orthogonally, \\ \therefore 2g(-2)+2g\left(-3\right)=10 \\ \Rightarrow -4g-6g=10 \\ \Rightarrow -10g=10 \\ \Rightarrow g=\frac{10}{-10}=-1 \\ Thus required equation of circle is x^2+y^2-2x-2y=0. \\ \left[If circle x^2+y^2+2gx+2hy+f=0 intersects circle x^2+y^2+2g_{1}x+2h_{1}y+f_1=0, then 2g{g}_1+2h{h}_1=f+f_1.\right] \end{gathered}$$
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