1. **Problem statement:** We have two circles and a tangent line. Circle P has center at (0, 2) and radius 10. Point A (6, 10) lies on circle P and the tangent at A touches circle P. Circle Q is centered where this tangent meets the y-axis, with radius $ \frac{5}{2} \sqrt{5} $. 2. **Equation of circle P:** The general equation of a circle with center $(h, k)$ and radius $r$ is: $$ (x - h)^2 + (y - k)^2 = r^2 $$ For circle P, $h=0$, $k=2$, and $r=10$, so: $$ (x - 0)^2 + (y - 2)^2 = 10^2 $$ $$ x^2 + (y - 2)^2 = 100 $$ 3. **Equation of the tangent to circle P at A:** The slope of the radius from center $(0, 2)$ to point A $(6, 10)$ is: $$ m_{radius} = \frac{10 - 2}{6 - 0} = \frac{8}{6} = \frac{4}{3} $$ The tangent is perpendicular to the radius, so its slope is the negative reciprocal: $$ m_{tangent} = -\frac{3}{4} $$ Using point-slope form with point A: $$ y - 10 = -\frac{3}{4}(x - 6) $$ Simplify: $$ y = -\frac{3}{4}x + \frac{18}{4} + 10 = -\frac{3}{4}x + \frac{9}{2} + 10 = -\frac{3}{4}x + \frac{29}{2} $$ 4. **Equation of circle Q and verifying y-coordinates of intersection points:** The tangent meets the y-axis where $x=0$: $$ y = -\frac{3}{4}(0) + \frac{29}{2} = \frac{29}{2} = 14.5 $$ So center of circle Q is $ (0, 14.5) $. Radius of Q is: $$ r_Q = \frac{5}{2} \sqrt{5} $$ Equation of circle Q: $$ (x - 0)^2 + (y - 14.5)^2 = \left( \frac{5}{2} \sqrt{5} \right)^2 = \frac{25}{4} \times 5 = \frac{125}{4} $$ To verify the y-coordinates of intersection points of circles P and Q are 11, substitute $y=11$ into both equations: For circle P: $$ x^2 + (11 - 2)^2 = 100 $$ $$ x^2 + 9^2 = 100 $$ $$ x^2 + 81 = 100 $$ $$ x^2 = 19 $$ For circle Q: $$ x^2 + (11 - 14.5)^2 = \frac{125}{4} $$ $$ x^2 + (-3.5)^2 = \frac{125}{4} $$ $$ x^2 + 12.25 = 31.25 $$ $$ x^2 = 19 $$ Since both give $x^2 = 19$, the points of intersection have y-coordinate 11. 5. **Coordinates of intersection points of tangent and circle Q:** Equation of tangent: $$ y = -\frac{3}{4}x + \frac{29}{2} $$ Equation of circle Q: $$ x^2 + (y - 14.5)^2 = \frac{125}{4} $$ Substitute $y$ from tangent into circle Q: $$ x^2 + \left(-\frac{3}{4}x + \frac{29}{2} - 14.5\right)^2 = \frac{125}{4} $$ Simplify inside the square: $$ -\frac{3}{4}x + \frac{29}{2} - \frac{29}{2} = -\frac{3}{4}x + 0 = -\frac{3}{4}x $$ So: $$ x^2 + \left(-\frac{3}{4}x\right)^2 = \frac{125}{4} $$ $$ x^2 + \frac{9}{16}x^2 = \frac{125}{4} $$ $$ \left(1 + \frac{9}{16}\right) x^2 = \frac{125}{4} $$ $$ \frac{25}{16} x^2 = \frac{125}{4} $$ Multiply both sides by $\frac{16}{25}$: $$ x^2 = \frac{125}{4} \times \frac{16}{25} = \frac{125 \times 16}{4 \times 25} = \frac{2000}{100} = 20 $$ $$ x = \pm \sqrt{20} = \pm 2\sqrt{5} $$ Find corresponding y values: $$ y = -\frac{3}{4} (\pm 2\sqrt{5}) + \frac{29}{2} = \mp \frac{3}{4} \times 2\sqrt{5} + \frac{29}{2} = \mp \frac{3}{2} \sqrt{5} + \frac{29}{2} $$ **Final answers:** - (a) Equation of circle P: $$ x^2 + (y - 2)^2 = 100 $$ - (b) Equation of tangent at A: $$ y = -\frac{3}{4}x + \frac{29}{2} $$ - (c) Equation of circle Q: $$ x^2 + (y - 14.5)^2 = \frac{125}{4} $$ with intersection y-coordinates verified as 11. - (d) Points of intersection of tangent and circle Q: $$ \left(2\sqrt{5}, \frac{29}{2} - \frac{3}{2} \sqrt{5}\right) \text{ and } \left(-2\sqrt{5}, \frac{29}{2} + \frac{3}{2} \sqrt{5}\right) $$