1. We are given the system of equations representing two circles: $$ (x-10)^2 + (y-5)^2 = 25 $$ $$ (x-20)^2 + (y-10)^2 = 100 $$ We need to find the points where these two circles intersect. 2. Expand both equations: $$ (x-10)^2 + (y-5)^2 = x^2 - 20x + 100 + y^2 - 10y + 25 = 25 $$ Simplify: $$ x^2 - 20x + y^2 - 10y + 125 = 25 $$ $$ x^2 + y^2 - 20x - 10y + 100 = 0 \, ...(1) $$ Similarly for the second circle, $$ (x-20)^2 + (y-10)^2 = x^2 - 40x + 400 + y^2 - 20y + 100 = 100 $$ Simplify: $$ x^2 + y^2 - 40x - 20y + 500 = 100 $$ $$ x^2 + y^2 - 40x - 20y + 400 = 0 \, ...(2) $$ 3. Subtract equation (1) from (2) to eliminate $x^2 + y^2$: $$ (x^2 + y^2 - 40x - 20y + 400) - (x^2 + y^2 - 20x - 10y + 100) = 0 $$ Simplify: $$ -40x - 20y + 400 + 20x + 10y - 100 = 0 $$ $$ (-40x + 20x) + (-20y + 10y) + (400 - 100) = 0 $$ $$ -20x - 10y + 300 = 0 $$ 4. Divide entire equation by -10: $$ 2x + y - 30 = 0 $$ Rearranged: $$ y = 30 - 2x \, ...(3) $$ 5. Substitute $y = 30 - 2x$ from (3) into equation (1): $$ x^2 + (30 - 2x)^2 - 20x - 10(30 - 2x) + 100 = 0 $$ Expand terms: $$ x^2 + (900 - 120x + 4x^2) - 20x - 300 + 20x + 100 = 0 $$ Simplify: $$ x^2 + 900 - 120x + 4x^2 - 20x - 300 + 20x + 100 = 0 $$ Combine like terms: $$ 5x^2 - 120x + (900 - 300 + 100) = 0 $$ $$ 5x^2 - 120x + 700 = 0 $$ 6. Divide entire equation by 5: $$ x^2 - 24x + 140 = 0 $$ 7. Solve quadratic equation using the quadratic formula: $$ x = \frac{24 \pm \sqrt{(-24)^2 - 4 \cdot 1 \cdot 140}}{2} $$ Calculate discriminant: $$ \sqrt{576 - 560} = \sqrt{16} = 4 $$ 8. Find the two solutions for $x$: $$ x_1 = \frac{24 + 4}{2} = \frac{28}{2} = 14 $$ $$ x_2 = \frac{24 - 4}{2} = \frac{20}{2} = 10 $$ 9. Find corresponding $y$ values by substituting back into $y = 30 - 2x$: For $x=14$: $$ y = 30 - 2 \cdot 14 = 30 - 28 = 2 $$ For $x=10$: $$ y = 30 - 2 \cdot 10 = 30 - 20 = 10 $$ 10. The points of intersection are: $$ (14, 2) \text{ and } (10, 10) $$