1. **State the problem:** We have points A(-1, 2) and B(11, 6) on a circle with center P(7, k). M is the midpoint of AB, and line l passes through M and P. 2. **Find midpoint M of AB:** Midpoint formula: $$M = \left(\frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2}\right)$$ Calculate: $$M_x = \frac{-1 + 11}{2} = \frac{10}{2} = 5$$ $$M_y = \frac{2 + 6}{2} = \frac{8}{2} = 4$$ So, $$M = (5, 4)$$ 3. **Find equation of line l through M(5,4) and P(7,k):** Slope formula: $$m = \frac{k - 4}{7 - 5} = \frac{k - 4}{2}$$ Equation in point-slope form: $$y - 4 = m(x - 5)$$ Substitute slope: $$y - 4 = \frac{k - 4}{2}(x - 5)$$ Rewrite in slope-intercept form: $$y = \frac{k - 4}{2}x - \frac{5(k - 4)}{2} + 4$$ Simplify constant term: $$y = \frac{k - 4}{2}x - \frac{5k - 20}{2} + 4 = \frac{k - 4}{2}x - \frac{5k}{2} + 10 + 4 = \frac{k - 4}{2}x - \frac{5k}{2} + 14$$ 4. **Find value of k using the fact that P is center of circle passing through A and B:** Distance PA = Distance PB (both equal radius) Calculate distance PA: $$PA = \sqrt{(7 + 1)^2 + (k - 2)^2} = \sqrt{8^2 + (k - 2)^2} = \sqrt{64 + (k - 2)^2}$$ Calculate distance PB: $$PB = \sqrt{(7 - 11)^2 + (k - 6)^2} = \sqrt{(-4)^2 + (k - 6)^2} = \sqrt{16 + (k - 6)^2}$$ Set equal: $$64 + (k - 2)^2 = 16 + (k - 6)^2$$ Expand squares: $$(k - 2)^2 = k^2 - 4k + 4$$ $$(k - 6)^2 = k^2 - 12k + 36$$ Substitute: $$64 + k^2 - 4k + 4 = 16 + k^2 - 12k + 36$$ Simplify: $$68 - 4k = 52 - 12k$$ Bring variables to one side: $$-4k + 12k = 52 - 68$$ $$8k = -16$$ $$k = -2$$ 5. **Equation of line l with k = -2:** Slope: $$m = \frac{-2 - 4}{2} = \frac{-6}{2} = -3$$ Intercept: $$c = - \frac{5k}{2} + 14 = - \frac{5(-2)}{2} + 14 = 5 + 14 = 19$$ Equation: $$y = -3x + 19$$ 6. **Equation of the circle:** Center: $$P(7, -2)$$ Radius: distance PA (or PB) with k = -2 $$r = \sqrt{(7 + 1)^2 + (-2 - 2)^2} = \sqrt{8^2 + (-4)^2} = \sqrt{64 + 16} = \sqrt{80} = 4\sqrt{5}$$ Circle equation: $$(x - 7)^2 + (y + 2)^2 = (4\sqrt{5})^2 = 80$$