Find the equation of the circle which passes through the points `(2,3),(4,2)` and the centre lies on the straight line `y-4x+3=0.`

Explanation

Let the general equation of the circle is
`x^(2)+y^(2)+2gx+2fy+c=0` ….(i)
Since, this circle passes through the points (2,3) and (4,5)
`therefore 4+9+4g+6f+c=0`
`rArr 4g+6f+c=-13`…(ii)
and `16+25+8g+10f+c=0`
`rArr 8g+10f+c=-41`....(iii)
Since, the centre of the circle `(-g,-f)` lies on the straight line `y-4x+3=0`
i.e., `+4g-f+3=0`...(iv)
From Eq. (iv) `4g=f-3`
On putting `4g=f-3` in Eq. (ii), we get
`f-3+6f+c=-13`
`rArr 7f+c=10`...(v)
From Eqs. (ii) and (iii),
`{:(8g+12f+2c=-26),(8g+10f+c=-41),("- " "- ""-"" "+),("___________________"),(" "2f+c=15):}`
From Eqs. (ii) and (vi),
`{:(7f+c=-10),(2f+c=15),("- " "- ""-"),("__________"),(5f=-25):}`
Now, `c=10+15=25`
From Eq. (iv) `4g+5+3=0`
. `rArrg=-2`
From Eq. (i), equation of the circle is `x^(2)+y^(2)-4x-10y+25=0`