Math  /  Geometry

QuestionFind the equation of the circle that passes through the points A(0,3),B(4,1)A(0,3), B(4,-1) and having its center on the line 3x+4y+1=03 x+4 y+1=0. [6 marks]

Studdy Solution

STEP 1

1. The circle passes through the points A(0,3) A(0, 3) and B(4,1) B(4, -1) .
2. The center of the circle lies on the line 3x+4y+1=0 3x + 4y + 1 = 0 .

STEP 2

1. Set up the general equation of a circle.
2. Use the condition that the circle passes through the given points to form equations.
3. Use the condition that the center lies on the given line to form another equation.
4. Solve the system of equations to find the center and radius.
5. Write the equation of the circle.

STEP 3

Set up the general equation of a circle with center (h,k)(h, k) and radius rr:
(xh)2+(yk)2=r2 (x - h)^2 + (y - k)^2 = r^2

STEP 4

Substitute the point A(0,3) A(0, 3) into the circle's equation:
(0h)2+(3k)2=r2 (0 - h)^2 + (3 - k)^2 = r^2 h2+(3k)2=r2 h^2 + (3 - k)^2 = r^2
Substitute the point B(4,1) B(4, -1) into the circle's equation:
(4h)2+(1k)2=r2 (4 - h)^2 + (-1 - k)^2 = r^2 (4h)2+(1k)2=r2 (4 - h)^2 + (-1 - k)^2 = r^2

STEP 5

Use the condition that the center (h,k)(h, k) lies on the line 3x+4y+1=0 3x + 4y + 1 = 0 :
3h+4k+1=0 3h + 4k + 1 = 0

STEP 6

Solve the system of equations:
1. h2+(3k)2=r2 h^2 + (3 - k)^2 = r^2
2. (4h)2+(1k)2=r2 (4 - h)^2 + (-1 - k)^2 = r^2
3. 3h+4k+1=0 3h + 4k + 1 = 0

First, express k k in terms of h h from the line equation:
k=3h+14 k = -\frac{3h + 1}{4}
Substitute k=3h+14 k = -\frac{3h + 1}{4} into the first two circle equations and solve for h h .
After substitution and simplification, solve for h h and k k .

STEP 7

(continued) Solve the equations obtained in STEP_4 to find the values of h h and k k .
Substitute back to find r2 r^2 .

STEP 8

Write the equation of the circle using the found values of h h , k k , and r r :
(xh)2+(yk)2=r2 (x - h)^2 + (y - k)^2 = r^2
The equation of the circle is:
(xh)2+(yk)2=r2 \boxed{(x - h)^2 + (y - k)^2 = r^2}

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