Find the equation of the circle passing through the point `(7, 3)` having radius `3` units and whose centre lies on the line `y = x-1`

Explanation

Let equation of circle be
`(x-h)^(2)+(y-k)^(2)=r^(2)…..(i)`
`rArr (x-h)^(2)+(y-k)^(2)=9`
Given that, centre (h,k) lies on the line
y=x-1 i.e., k=h-1
Now, the circle passes through the point (7,3).
`(7-h)^(2)+(3-k)^(3)=9`
`rArr 49-14h+h^(2)+9-6k+k^(2)=9`
`rArr h^(2)+k^(2)-14h-6k+49=0`.....(iii)
On putting k=h-1 in Eq. (iii) we get
`h^(2)+(h-1)^(2)-14h-6(h-1)+49=0`
`rArr 2h^(2)-22h+56=0`
`rArrh^(2)-11h+28=0`
`rArr h^(2)-7h-4h+28=0`
`rArrh(h-7)-4(h-7)=0`
`rArr(h-7)(h-4)=0`
`therefore h=4,7`
When h=7, then k=7-1=6
`therefore` Centre(7,6)
When h=4, then k=3
`therefore` Centre-(4,3)
So, the equation of circle when centre (7,6) is
`(x-7)^(2)+(y-6)^(2)=9`
`rArr x^(2)-14x+49+y^(2)-12y+36=9`
`rArr x^(2)+y^(2)-14x-12y+76=0`
When centre (4,3), then the equation of the circle is
`(x-4)^(2)+(y-3)^(2)=9`
`therefore x^(2)-8x+16+y^(2)-6y+9=9`
`rArr x^(2)+^(2)-8x-6y+16=0`