The equation of chord of the circle

Question

x² + y² -2x - 3=0 whose midpoint is (1, 1) is

Answer 1

The given circle x^2 + y^2 –2x–3=0 can be rewritten as

x^2– 2x + 1 +y^2 = 4, or (x–1)^2 +y^2 =4, which has center (1,0) and radius 2.

Let’s find out where it intersects the line x+y=3, in order to find the endpoints of the chord. We will substitute y=-x+3 into the equation of the circle,

(x–1)^2 + (-x+3)^2=4

x^2–2x+1+x^2–6x+9=4

2x^2–8x+6=0

x^2–4x+3=0

(x–3)(x–1)=0

x=3, so y=-x+3, y=0 (3,0)

x=1 so y=-x+3, y=2 (1,2)

So we are looking for the equation of a circle with diameter having endpoints (3,0) and (1,2). The length of this diameter is √[(3–1)^2+(0–2)^2], which is √8=2√2, so the length of the radius of the circle would be √2.

The midpoint of the diameter of this circle gives us its center. Using the midpoint formula, [(3+1)/2, (0+2)/2], we find the center is (2,1).

The equation of a circle with center (2,1) and radius √2 is

(x–2)^2+(y–1)^2=2

Answer 2

Complete the square to find the centre of the circle
x^2 - 2x + 1 + y^2 = 3+1
(x-1)^2 + y^2 = 4

centre is (1,0), radius is 2

make a rough sketch to see that the line from the centre(1,0) to the midpoint (1,1) of the chord is a vertical line x = 1
So the chord, which is perpendicular to the line from (1,0) to (1,1) must
have equation y = 1

sub in y = 1 into the circle:
(x-1)^2 + 1 = 4
(x-1)^2 = 3
x-1 = ± √3
x = 1 ± √3

Just realized that all you wanted was the equation of the chord, which was
y = 1

Answer 3

Question for Tiger:
Where does the line x+y=3 come from, and what does it have to to with this?