1. **Problem statement:** We need to find the equation of a circle with radius 6 and then find a chord passing through the intersection points of this circle and another circle. 2. **Circle equation:** The general equation of a circle with center $(h,k)$ and radius $r$ is: $$ (x - h)^2 + (y - k)^2 = r^2 $$ 3. **Assuming the new circle is centered at the origin** for simplicity, its equation is: $$ x^2 + y^2 = 6^2 = 36 $$ 4. **Intersection of two circles:** Suppose the other circle has equation: $$ (x - h_1)^2 + (y - k_1)^2 = r_1^2 $$ The intersection points satisfy both circle equations simultaneously. 5. **Chord through intersection points:** The line passing through the intersection points of two circles is called the radical line. Its equation can be found by subtracting the two circle equations: $$ (x - h)^2 + (y - k)^2 - (x - h_1)^2 - (y - k_1)^2 = r^2 - r_1^2 $$ 6. **Example:** Let the other circle be centered at $(3,0)$ with radius 5: $$ (x - 3)^2 + y^2 = 25 $$ Subtracting the two equations: $$ x^2 + y^2 - ((x - 3)^2 + y^2) = 36 - 25 $$ $$ x^2 - (x^2 - 6x + 9) = 11 $$ $$ x^2 - x^2 + 6x - 9 = 11 $$ $$ 6x = 20 $$ $$ x = \frac{10}{3} $$ 7. **Chord equation:** The chord passing through the intersection points is the vertical line: $$ x = \frac{10}{3} $$ **Final answer:** - Circle equation: $$ x^2 + y^2 = 36 $$ - Chord passing through intersection points: $$ x = \frac{10}{3} $$