determine the general form of the equation of the circle whose center is (3,2) and whose graph contains the point (1,2)

Question

sketch the graph

Answer 1

determine the general form of the equation of the circle whose center is (3,2) and whose graph contains the point (1,2)

The general eqaution of a circle, centred on the point (a, b) and of radius = R, is

(x-a)^2 + (y-b)^2 = R^2

We are given that the centre is O(3,2), so we can write the eqn of the circle as,

(x-3)^2 + (y-2)^2 = R^2

We are given that a point on the circle is P(1,2)

Using Pythagoras, the distance OP is given by sqrt{(Ox - Px)^2 + (Oy - Py)^2}

OP = sqrt{(3 - 1)^2 + (2 - 2)^2}

OP = sqrt{(2)^2}

OP = 2, i,e,

R = 2

The general equation of the circle now is,

(x - 3)^2 + (y - 2)^2 = 4