1. **Problem statement:** A circle passes through the points $(2, 3)$ and $(-1, 6)$, and its centre lies on the line $2x + 5y = -1$. We need to find: (i) the coordinates of the centre of the circle, (ii) the radius of the circle, (iii) the equation of the circle. 2. **Formula and approach:** The general equation of a circle with centre $(h, k)$ and radius $r$ is: $$ (x - h)^2 + (y - k)^2 = r^2 $$ Since the circle passes through points $(2, 3)$ and $(-1, 6)$, these points satisfy the equation: $$ (2 - h)^2 + (3 - k)^2 = r^2 $$ $$ (-1 - h)^2 + (6 - k)^2 = r^2 $$ Also, the centre $(h, k)$ lies on the line: $$ 2h + 5k = -1 $$ 3. **Step 1: Equate the distances from centre to the two points** Since both points lie on the circle, their distances to the centre are equal: $$ (2 - h)^2 + (3 - k)^2 = (-1 - h)^2 + (6 - k)^2 $$ Expanding both sides: $$ (2 - h)^2 = (2 - h)(2 - h) = 4 - 4h + h^2 $$ $$ (3 - k)^2 = 9 - 6k + k^2 $$ $$ (-1 - h)^2 = ( -1 - h)( -1 - h) = 1 + 2h + h^2 $$ $$ (6 - k)^2 = 36 - 12k + k^2 $$ Substitute: $$ 4 - 4h + h^2 + 9 - 6k + k^2 = 1 + 2h + h^2 + 36 - 12k + k^2 $$ Simplify by canceling $h^2$ and $k^2$ on both sides: $$ 13 - 4h - 6k = 37 + 2h - 12k $$ Bring all terms to one side: $$ 13 - 4h - 6k - 37 - 2h + 12k = 0 $$ Simplify: $$ (13 - 37) + (-4h - 2h) + (-6k + 12k) = 0 $$ $$ -24 - 6h + 6k = 0 $$ Divide entire equation by 6: $$ rac{-24}{6} + rac{-6h}{6} + rac{6k}{6} = 0 $$ $$ -4 - h + k = 0 $$ Rewrite: $$ k = h + 4 $$ 4. **Step 2: Use the line equation for centre:** Given: $$ 2h + 5k = -1 $$ Substitute $k = h + 4$: $$ 2h + 5(h + 4) = -1 $$ Expand: $$ 2h + 5h + 20 = -1 $$ Simplify: $$ 7h + 20 = -1 $$ $$ 7h = -21 $$ $$ h = -3 $$ Then, $$ k = h + 4 = -3 + 4 = 1 $$ So, the centre is at $(-3, 1)$. 5. **Step 3: Find the radius $r$** Use the distance formula from centre to one of the points, say $(2, 3)$: $$ r = \sqrt{(2 - (-3))^2 + (3 - 1)^2} = \sqrt{(5)^2 + (2)^2} = \sqrt{25 + 4} = \sqrt{29} $$ 6. **Step 4: Write the equation of the circle** Using centre $(-3, 1)$ and radius $\sqrt{29}$: $$ (x + 3)^2 + (y - 1)^2 = 29 $$ **Final answers:** (i) Centre: $(-3, 1)$ (ii) Radius: $\sqrt{29}$ (iii) Equation: $$(x + 3)^2 + (y - 1)^2 = 29$$