Find the equation of the circle passing through the points `(2," "3)` and `(1," "1)` and whose centre is on the line `x" "" "3y" "" "11" "=" "0` .

Explanation

Let equation of circle be
`x^(2)+y^(2)+2gx+2fy+c=0`
This circle passes through the point (2,3).
`:.2^(2)+3^(2)+4g+6f+c=0`
`rArr" "4g+6f+c=-13` . . . (1)
This circle passes through the point (-1,1).
`:." "(-1)^(2)+(1)^(2)-2g+2f+c=0`
`rArr" "-2g+2f+c=-2` . . .(2)
The centre (-g,-f) of circle lies on the line
x-3y-11=0
`:." "-g+3g-11=0`
`rArr" "-g+3f=11` . . .(3)
Subtracting equation (2) from equation (1)
`6g+4f=-11` . . .(4)
Multiplying equation (3) by 6 and adding in equation (4)
`{:(-6g+18f=66),(ul(6g+4f=-11)),(" "22f=55):}`
`rArr" "f=(5)/(2)`
Put the value of f in equation (3)
`-g+(15)/(2)=11`
`rArr" "g=-(7)/(2)`
Put the value of g and is equation (1)
-14+15+c=-13
`rArr" "c=-14`
Therefore equation of circle is
`x^(2)+y^(2)-7x+5y-14=0`