1. **Problem statement:** We are given points $P(1,1)$ and $Q(7,11)$ which form the diameter of a circle. (a) Find the equation of the circle. (b) Find the equation of the tangent to the circle at point $Q$. --- 2. **Find the center and radius of the circle:** The center $C$ of the circle is the midpoint of the diameter $PQ$. Midpoint formula: $$C = \left(\frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2}\right)$$ Calculate: $$C = \left(\frac{1 + 7}{2}, \frac{1 + 11}{2}\right) = (4, 6)$$ Radius $r$ is half the length of $PQ$. Distance formula: $$d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$$ Calculate length $PQ$: $$d = \sqrt{(7 - 1)^2 + (11 - 1)^2} = \sqrt{6^2 + 10^2} = \sqrt{36 + 100} = \sqrt{136}$$ Radius: $$r = \frac{d}{2} = \frac{\sqrt{136}}{2} = \sqrt{34}$$ --- 3. **Equation of the circle:** The general form of a circle with center $(h,k)$ and radius $r$ is: $$ (x - h)^2 + (y - k)^2 = r^2 $$ Substitute $h=4$, $k=6$, and $r=\sqrt{34}$: $$ (x - 4)^2 + (y - 6)^2 = 34 $$ --- 4. **Find the equation of the tangent at point $Q(7,11)$:** The tangent line at $Q$ is perpendicular to the radius $CQ$. Calculate slope of radius $CQ$: $$m_{CQ} = \frac{11 - 6}{7 - 4} = \frac{5}{3}$$ Slope of tangent $m_t$ is the negative reciprocal: $$m_t = -\frac{3}{5}$$ Use point-slope form for tangent line at $Q(7,11)$: $$ y - 11 = m_t (x - 7) $$ Substitute $m_t$: $$ y - 11 = -\frac{3}{5}(x - 7) $$ Simplify: $$ y = -\frac{3}{5}x + \frac{21}{5} + 11 = -\frac{3}{5}x + \frac{21}{5} + \frac{55}{5} = -\frac{3}{5}x + \frac{76}{5} $$ --- **Final answers:** (a) Equation of the circle: $$ (x - 4)^2 + (y - 6)^2 = 34 $$ (b) Equation of the tangent line at $Q$: $$ y = -\frac{3}{5}x + \frac{76}{5} $$