Question

The centre of the circle \( (x-3)(x+1)+(y-1)(y+3)=0 \) is:

• A. \( (3, 1) \)
• B. \( (-1, -3) \)
• C. \( (3, -3) \)
• D. \( (-1, 1) \)
• E. \( (1, -1) \)

Hint:

To find the center of a circle from the equation \( (x-h)(x+k)+(y-p)(y+q)=0 \), expand the terms and simplify the equation into the standard form \( (x-h)^2 + (y-p)^2 = r^2 \).

Answer 1

Answer: (E)

We are given the equation of the circle in the form: \[ (x - 3)(x + 1) + (y - 1)(y + 3) = 0 \] Let's expand the terms in the equation:
1. For the first part: \[ (x - 3)(x + 1) = x^2 + x - 3x - 3 = x^2 - 2x - 3 \] 2. For the second part: \[ (y - 1)(y + 3) = y^2 + 3y - y - 3 = y^2 + 2y - 3 \] Substituting these into the original equation: \[ x^2 - 2x - 3 + y^2 + 2y - 3 = 0 \] Simplify the equation: \[ x^2 + y^2 - 2x + 2y - 6 = 0 \] Now, we rearrange the equation into standard form for a circle: \[ x^2 - 2x + y^2 + 2y = 6 \] Next, we complete the square for both \( x \) and \( y \).
- For \( x^2 - 2x \), take half of -2, which is -1, square it to get 1. So, add and subtract 1.
- For \( y^2 + 2y \), take half of 2, which is 1, square it to get 1. So, add and subtract 1.
Thus, we have: \[ (x^2 - 2x + 1) + (y^2 + 2y + 1) = 6 + 1 + 1 \] Simplifying: \[ (x - 1)^2 + (y + 1)^2 = 8 \] This is the standard equation of a circle with center \( (h, k) \) and radius \( \sqrt{8} \).
So, the center of the circle is \( (1, -1) \).
Thus, the correct answer is option (E), \( (1, -1) \).