According to figure – it is clear that **centre** will be ( h , h ) **radius** will be equal to **h**.

**( x – h )**^{2 }+ ( y – h )^{2} = h^{2}

**x**^{2 }+ **y**^{2 }— 2.h.x — 2.k.y + h^{2 }= 0

**But the centre of the circle could be any of the four quadrants. Then co-ordinates of the centre taken as**

** ( ± h , ± h ).**

**Then equation of the circle is —**

**x**^{2 }**+**** y**^{2 }**± **** 2.h.x ****± **** 2.k.y ****+ **** h**^{2 }= 0