Question

The equation of chord of the circle

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?

Answer 4

[AI-generated answer — Explain Bot]

To find the equation of the chord of a circle with a given midpoint, we need to follow these steps:

Step 1: Find the coordinates of the center of the circle
The equation of a circle is in the form (x - h)² + (y - k)² = r², where (h, k) represents the coordinates of the center of the circle. To get the center coordinates, we need to rewrite the given equation: x² + y² - 2x - 3 = 0.

Rearrange the equation to isolate the terms involving x and y:
x² - 2x + y² - 3 = 0.

Next, complete the square for both x and y terms:
(x² - 2x + 1) + (y² - 3) = 1.

Simplify further:
(x - 1)² + (y - 0)² = 4.

Comparing this equation to the general equation of a circle, we can determine that the center is at (1, 0).

Step 2: Find the radius of the circle
We can see from the equation (x - 1)² + (y - 0)² = 4 that the radius squared is 4. Thus, the radius is 2.

Step 3: Find the equation of the chord
Given that the midpoint of the chord is (1, 1), we can use this information to find the endpoints of the chord. Since the midpoint (1, 1) lies on the chord, the distance from the midpoint to the center of the circle should be equal to the radius.

Using the distance formula, we can calculate the distance between the midpoint (1, 1) and the center (1, 0):
√[(x2 - x1)² + (y2 - y1)²] = √[(1 - 1)² + (0 - 1)²] = √[0² + (-1)²] = √1 = 1.

Since this distance is not equal to the radius (2), the chord does not pass through the center of the circle. This means we can have multiple possible equations for the chord.

Therefore, without any additional information, we cannot uniquely determine the equation of the chord.