Two circles have the same radius.
The centre of one circle is (−2, 1).
The centre of the other circle is (3, −2).
The circles intersect at two distinct points.
What is the equation of the straight line through the two points at which the circles intersect?
- A3x − 5y = 4
- B3x + 5y = −1
- C5x − 3y = −4
- D5x − 3y = −1
- E5x − 3y = 1
- F5x − 3y = 4
- G5x + 3y = 1
Show the answer and worked solution
answer · F
- A3x − 5y = 4
- B3x + 5y = −1
- C5x − 3y = −4
- D5x − 3y = −1
- E5x − 3y = 1
- F5x − 3y = 4
- G5x + 3y = 1
Call the common radius r. The two circles are (x+2)2 + (y−1)2 = r2 and (x−3)2 + (y+2)2 = r2. A point lies on both exactly when it satisfies the difference of the two equations, and subtracting kills both x2 and y2 along with r2. That leaves 4x + 5 − 2y = −6x + 13 + 4y, so 10x − 6y = 8, that is 5x − 3y = 4. Notice the radius never had to be found: the common chord's line does not depend on it.