Find the equation of the circle …

The equation of the circle is, (x – h)2 + (y – k)2 = r2 ….(i) Since the circle passes through point (2, 3) ∴ (2 – h)2 + (3 – k)2 = r2 ⇒ 4 + h2 – 4h + 9 + k2 – 6k = r2 ⇒ h2+ k2 – 4h – 6k + 13 = r2 ….(ii) Also, the circle passes through point (-1, 1) ∴ (-1 – h)2 + (1 – k)2 = r2 ⇒ 1 + h2 + 2h + 1 + k2 – 2k = r2 ⇒ h2 + k2 + 2h – 2k + 2 = r2 ….(iii) From (ii) and (iii), we have h2 + k2 – 4h – 6k + 13 = h2 + k2 + 2h – 2k + 2 ⇒ -6h – 4k = -11 ⇒ 6h + 4k = 11 …(iv) Since the centre (h, k) of the circle lies on the line x – 3y-11 = 0. ∴ h – 3k – 11 = 0 ⇒ h -3k = 11 …(v) Solving (iv) and (v), we get h = \frac { 7 }{ 2 } and k = \frac { -5 }{ 2 } Putting these values of h and k in (ii), we get \left( \frac { 7 }{ 2 } \right) ^{ 2 }+\left( \frac { -5 }{ 2 } \right) ^{ 2 }-\frac { 4\times 7 }{ 2 } -6\times \frac { -5 }{ 2 } +13={ r }^{ 2 } ⇒ \frac { 49 }{ 4 } +\frac { 25 }{ 4 } -14+15+13 ⇒ { r }^{ 2 }=\frac { 65 }{ 2 } Thus required equation of circle is ⇒ \left( x-\frac { 7 }{ 2 } \right) ^{ 2 }+\left( y+\frac { 5 }{ 2 } \right) ^{ 2 }=\frac { 65 }{ 2 } ⇒ { x }^{ 2 }+\frac { 49 }{ 4 } -7x+{ y }^{ 2 }+\frac { 25 }{ 4 } +5y=\frac { 65 }{ 2 } ⇒ 4x2 + 49 – 28x + 4y2 + 25 + 20y = 130 ⇒ 4x2 + 4y2 – 28x + 20y – 56 = 0 ⇒ 4(x2 + y2 – 7x + 5y -14) = 0 ⇒ x2 + y2 – 7x + 5y -14 = 0.