Given two points B (3; 4) A (2; 1) a) Make an equation with the center at point A and passing through point B

Given two points B (3; 4) A (2; 1) a) Make an equation with the center at point A and passing through point B, b) Find the coordinates of the points of intersection with the obziss and ordinate axes

a) (x – a) ^ 2 + (y – b) ^ 2 = R ^ 2 – general form of the equation of the circle.

R = AB.

AB ^ 2 = (3 – 2) ^ 2 + (4 – 1) 2 = 10.

(x – 2) ^ 2 + (y – 1) ^ 2 = 10 – the required equation of the circle.

b) y = 0;

(x – 2) ^ 2 + (0 – 1) 2 = 10;

(x – 2) ^ 2 = 9;

x – 2 = 3 or x – 2 = – 3

x = 5; x = – 1.

(5; 0) and (- 1; 0) are the points of intersection of the circle with the abscissa axis.

x = 0;

(0 – 2) ^ 2 + (y – 1) ^ 2 = 10;

(y – 1) ^ 2 = 6;

y – 1 = √6 or y – 1 = – √6;

y = 1 + √6 or y = 1 – √6.

(0; 1 + √6); (0; 1 – √6) – points of intersection of the circle with the ordinate axis.