The circle (x − 3)^2 + (y − 2)^2 = 20 has centre C. i. Write down the radius of the circle and the coordinates of C. ii. Find the coordinates of the intersections of the circle with the x- and y-axes

The equation of the circle given by (x − 3)² + (y − 2)² = 20 indicates that the center of the circle, denoted as point C, is located at the coordinates (3, 2). To determine the radius of the circle, one can observe that the equation is in the standard form of a circle, where the radius is the square root of the constant on the right side of the equation. Therefore, the radius can be calculated as √20, which simplifies to 2√5.

To find the points where the circle intersects the x-axis, one must set y equal to zero in the equation of the circle. Substituting y = 0 into the equation yields (x − 3)² + (0 − 2)² = 20. This simplifies to (x − 3)² + 4 = 20, leading to (x − 3)² = 16. Taking the square root of both sides results in two possible values for x: x = 3 + 4 and x = 3 – 4, which gives the intersection points (7, 0) and (-1, 0) on the x-axis.

Similarly, to find the intersection points with the y-axis, one sets x equal to zero in the original equation. By substituting x = 0, the equation becomes (0 − 3)² + (y − 2)² = 20, which simplifies to 9 + (y − 2)² = 20. This leads to (y − 2)² = 11, resulting in two values for y: y = 2 + √11 and y = 2 – √11. Consequently, the intersection points on the y-axis are (0, 2 + √11) and (0, 2 – √11). For those seeking further assistance in understanding these concepts, a mathematics tutor can provide valuable guidance.